Light – Reflection and Refraction - Class 10 Science - Chapter 9 - Notes, NCERT Solutions & Extra Questions
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Notes - Light – Reflection and Refraction | Class 10 NCERT | Science
Comprehensive Class 10 Notes on Light Reflection and Refraction
Light plays a crucial role in our daily lives, allowing us to see the world around us. This article delves into the concepts of light reflection and refraction, essential topics for Class 10 students. Whether you're preparing for exams or just seeking to understand these phenomena, this comprehensive guide covers all necessary details while optimizing for the keyword "light reflection and refraction class 10 notes."
Reflection of Light
Laws of Reflection
Reflection occurs when light bounces off a surface. The fundamental laws of reflection are:
- The angle of incidence is equal to the angle of reflection.
- The incident ray, the normal to the mirror at the point of incidence, and the reflected ray all lie in the same plane.
Image Formation using Plane Mirrors
Plane mirrors are flat surfaces that reflect light. The images formed by plane mirrors possess the following properties:
- Virtual (cannot be captured on a screen)
- Erect (upright)
- Same size as the object
- Laterally inverted (left-right reversal)
Spherical Mirrors
Spherical mirrors can be either concave (curved inward) or convex (curved outward). They follow the laws of reflection but can form real or virtual images depending on the object's position.
Important Terms in Spherical Mirrors
- Pole (P): The center of the mirror's surface.
- Center of Curvature (C): The center of the sphere from which the mirror segment is taken.
- Radius of Curvature (R): The radius of the sphere.
- Principal Axis: The line passing through the pole and the center of curvature.
- Principal Focus (F): The point where parallel light rays meet (concave) or appear to diverge from (convex).
Image Formation with Concave Mirrors
The nature and size of the image formed by a concave mirror vary with the object's position:
- Beyond C: Real, inverted, smaller.
- At C: Real, inverted, same size.
- Between F and C: Real, inverted, larger.
- At F: No clear image (light rays parallel).
- Between F and P: Virtual, erect, larger.
Image Formation with Convex Mirrors
Convex mirrors always produce virtual, erect, and diminished images, irrespective of the object's position.
Mirror Formula and Magnification
The mirror formula is (\frac{1}{f} = \frac{1}{v} + \frac{1}{u}), where:
- (f) is the focal length
- (v) is the image distance
- (u) is the object distance
Refraction of Light
Refraction involves the bending of light as it passes from one medium to another due to a change in speed.
Understanding Refraction
Common phenomena like a pencil appearing bent in water occur due to refraction. The extent of bending depends on the media's refractive indices.
Laws of Refraction (Snell's Law)
Refraction follows Snell's Law:
- The incident ray, the refracted ray, and the normal to the interface all lie in the same plane.
- (\frac{\sin i}{\sin r} = \text{constant} = n), where (i) is the angle of incidence, (r) is the angle of refraction, and (n) is the refractive index.
Refraction Through a Glass Slab
A glass slab causes light to bend toward the normal when entering and away from the normal when exiting, resulting in a lateral shift but no change in direction of the emergent ray.
Spherical Lenses
Lenses are transparent materials with at least one curved surface and are categorized as convex (converging) or concave (diverging).
Types of Lenses
- Convex Lenses: Thicker at the center and converge light rays.
- Concave Lenses: Thinner at the center and diverge light rays.
Important Terms in Lenses
- Optical Centre (O): The central point of the lens.
- Principal Axis: The line passing through the optical center and the centers of curvature.
- Principal Focus (F): The point where parallel rays converge (convex) or appear to diverge from (concave).
Image Formation by Convex Lenses
Depending on the object's position:
- Beyond 2F: Real, inverted, smaller.
- At 2F: Real, inverted, same size.
- Between F and 2F: Real, inverted, larger.
- At F: No image.
- Between F and O: Virtual, erect, larger.
Image Formation by Concave Lenses
Concave lenses always produce virtual, erect, and diminished images.
Lens Formula and Magnification
The lens formula is (\frac{1}{f} = \frac{1}{v} - \frac{1}{u}). Magnification ((m)) is given by (\frac{h'}{h} = \frac{v}{u}), where (h') is the image height and (h) is the object height.
Power of a Lens
The power (P) of a lens is ( P = \frac{1}{f}) (in meters). It's positive for convex lenses and negative for concave lenses.
Applications of Reflection and Refraction
- Mirrors: Used in household items, vehicles, telescopes, and more.
- Lenses: Found in spectacles, magnifiers, cameras, microscopes, and telescopes.
Conclusion
Understanding the principles of light reflection and refraction is fundamental to studying various optical phenomena and their applications in technology and daily life.
By mastering these concepts, students can excel in their studies and appreciate the role of light in the natural and technological world.
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Extra Questions - Light – Reflection and Refraction | NCERT | Science | Class 10
Can absolute refractive index be less than 1? Justify.
The refractive index of a medium is always greater than or equal to 1; it cannot be less than 1.
The reason lies in the definition of refractive index ($ n $) of a medium (e.g., glass). It is defined as the ratio of the speed of light in a vacuum ($ c $) to the speed of light in the medium ($ v $):
$$ n = \frac{c}{v} $$
The minimum value of $ n $ can be 1, which is the refractive index of a vacuum or air (as air's refractive index is approximately the same as that of a vacuum). A refractive index below 1 implies that the speed of light in that medium is greater than the speed of light in vacuum, which is not possible in reality. Thus, a refractive index below 1 is simply not possible.
Lateral displacement of light by a glass slab having thickness 10 m is t. Then the lateral displacement by a glass slab having thickness 20 m will be:
Option 1) greater than t
Option 2) less than t
Option 3) equal to t
The correct option is 1) greater than t.
Explanation:
The lateral displacement of light in a glass slab is directly proportional to the thickness of the slab. As the thickness increases, the lateral displacement also increases. Therefore, when the thickness of the glass slab increases from 10 m to 20 m, the lateral displacement will be greater than t.
An object is placed in front of a convex lens, find the linear magnification produced by the lens.
A. 4
B. 3
C. 2
D. 1
The correct option is A: 4
To determine the linear magnification produced by a convex lens, we can use the lens formula:
$$ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} $$
Given:
Object distance, $u = -30 \text{ cm}$
Focal length of the lens, $f = 40 \text{ cm}$
Substitute into the lens formula:
$$ \frac{1}{v} + \frac{1}{30} = \frac{1}{40} $$
Rearrange the equation to solve for $\frac{1}{v}$:
$$ \frac{1}{v} = \frac{1}{40} - \frac{1}{30} $$
Find a common denominator:
$$ \frac{1}{v} = \frac{3}{120} - \frac{4}{120} = \frac{-1}{120} $$
Thus, the image distance $v$ is:
$$ v = -120 \text{ cm} $$
The linear magnification $m$ is given by the ratio of the image distance $v$ to the object distance $u$:
$$ m = \frac{v}{u} = \frac{-120}{-30} = +4 $$
Therefore, the linear magnification produced by the convex lens is 4.
The magnification of a concave mirror is, when the object is placed between the focus and the pole of the mirror.
A. negative
B. positive
C. zero
The correct option is B: Positive.
When an object is placed between the focus (F) and the pole (P) of a concave mirror, a virtual, erect, and enlarged (magnified) image is formed behind the mirror.
Since the height of the image ($ h_{i} $) is measured upwards from the principal axis, the image height is positive in this scenario. The object height ($ h_{o} $) is always considered positive for any type of spherical mirror.
Given the formula for magnification:
$$ m = \frac{h_{i}}{h_{o}} $$
where:
$ h_{i} $ is the height of the image,
$ h_{o} $ is the height of the object,
the magnification ($ m $) is thus positive.
What is Tyndall effect? Do true solutions exhibit the Tyndall effect?
The Tyndall effect refers to the phenomenon where colloidal particles scatter light. When a beam of light passes through a colloid, the particles within the colloid scatter the light, making the path of the light beam visible. This scattering is significant enough to be observed because the size of the colloidal particles is comparable to the wavelength of the light.
True solutions do not exhibit the Tyndall effect. This is because the particles in true solutions are too small to scatter light effectively. Only colloidal and suspension particles have the appropriate size to cause noticeable scattering of light, thus showing the Tyndall effect.
How is the focal length of a spherical mirror related to the radius of curvature?
The focal length ($f$) of a spherical mirror is directly related to its radius of curvature ($R$). This relationship can be expressed mathematically as:
$$ f = \frac{R}{2} $$
In other words, the focal length of a spherical mirror is half of its radius of curvature.
A concave mirror forms an image of 20 cm high object on a screen placed 5.0 m away from the mirror. The height of the image is 50 cm. Find the focal length of the mirror and the distance between the mirror and the object.
Given:
Height of the object, $ \mathrm{h_1} = 20 \text{ cm} $
Distance of the image from the mirror, $ v = -5 \text{ m} = -500 \text{ cm} $ (since the image is real and inverted)
Height of the image, $ \mathrm{h_2} = 50 \text{ cm} $
Using the magnification formula for a concave mirror: $$ \frac{-v}{u} = \frac{h_2}{h_1} $$
Substituting the given values: $$ \frac{500}{u} = \frac{50}{20} $$ $$ \frac{500}{u} = 2.5 $$
Rearranging to find $u$ (the object distance): $$ u = \frac{500}{2.5} $$ $$ u = -200 \text{ cm} = -2 \text{ m} $$
Now, using the mirror equation to find the focal length $f$: $$ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $$
Substituting the values for $v$ and $u$: $$ \frac{1}{-500} + \frac{1}{-200} = \frac{1}{f} $$ $$ -\frac{1}{500} - \frac{1}{200} = \frac{1}{f} $$ $$ \frac{1}{f} = -\frac{2}{1000} - \frac{5}{1000} $$ $$ \frac{1}{f} = -\frac{7}{1000} $$ $$ f = -\frac{1000}{7} $$ $$ f = -142.86 \text{ cm} $$ Converting to meters: $$ f = -1.44 \text{ m} $$
Therefore, the focal length of the concave mirror is ** $1.44 \text{ m}$ ** and the distance between the mirror and the object is ** $2 \text{ m}$ .
What do you mean by reflection of light?
Reflection of light refers to the phenomenon where light bounces back into the same medium after hitting a surface. When light encounters a surface, it does not pass through but is instead reflected, following the principle that the angle of incidence equals the angle of reflection. This behavior is crucial in various applications such as mirrors, optical instruments, and even natural observations like the way we see objects around us.
The image formed by the plane mirror is virtual and erect.
The correct option is C: virtual and erect.
A plane mirror forms an image that is virtual and erect. This means that the image appears to be located behind the mirror and is oriented in the same direction as the object. Additionally, the image is of the same size as the object. It is important to note that virtual images cannot be projected onto a screen because they do not actually exist at a specific location in space; they only appear to be behind the mirror.
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Ask Chatterbot AINCERT Solutions - Light – Reflection and Refraction | NCERT | Science | Class 10
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
The correct answer is (d) Clay.
Lenses work by refracting (bending) light, which requires the material to be transparent so that light can pass through it and be focused. Water, Glass, and Plastic are all materials that can be transparent and have the ability to refract light, making them suitable for making lenses. On the other hand, Clay is opaque and does not allow light to pass through, making it unsuitable for making lenses.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
The image formed by a concave mirror can be virtual, erect, and larger than the object only when the object is placed between the pole of the mirror and its principal focus. This is because when the object is placed in this position, the light rays from the object reflect off the mirror and diverge. Since the reflected rays diverge, they appear to come from a point behind the mirror when traced backward. The image formed in this case is virtual (because it cannot be obtained on a screen), erect (upright), and magnified (larger than the object).
So, the correct answer is:
(d) Between the pole of the mirror and its principal focus.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
For a convex lens, to obtain a real image of the same size as the object, the object should be placed at twice the focal length from the lens. This position is also known as "2F" or the second principal focus. When placed at this distance, the image formed is real, inverted, and of the same size as the object. It is also formed on the opposite side of the lens at the same distance from the lens as the object.
Hence, the correct answer is:
(b) At twice the focal length
A spherical mirror and a thin spherical lens have each a focal length of $-15 \mathrm{~cm}$. The mirror and the lens are likely to be
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
In optics, the sign convention for the focal length of mirrors and lenses is different based on their types:
For mirrors:
A convex mirror always has a positive focal length because the focus (F) is behind the mirror.
A concave mirror always has a negative focal length because the focus (F) is in front of the mirror.
For lenses:
A convex lens (also known as a converging lens) has a positive focal length because it converges light rays to a point.
A concave lens (also known as a diverging lens) has a negative focal length because it diverges light rays away from a point.
Given that both the mirror and the lens have a focal length of (-15 , \text{cm}), which is negative:
The mirror, having a negative focal length, is concave.
The lens, also having a negative focal length, is concave.
Therefore, the correct option is:
(a) both concave.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.
When you look into a mirror and regardless of how far you stand from it, if your image always appears erect, the mirror is likely to be either a plane mirror or a convex mirror.
In a plane mirror, images are always virtual, erect, and of the same size as the object. The distance from the object to the mirror is the same as the distance from the image to the mirror.
In a convex mirror, images are also virtual and erect, but they appear smaller than the object. Convex mirrors have a wider field of view compared to plane mirrors, which is why they are often used for security and rear-view mirrors on vehicles.
On the other hand, a concave mirror can produce both real and virtual images depending on the position of the object relative to the focus and the center of curvature of the mirror. When used to view images at a distance closer than the focus, concave mirrors can produce magnified, erect, and virtual images, but in general circumstance and especially at a distance, they are more known for real, inverted, and possibly magnified or diminished images depending on the specific positioning.
Therefore, the correct answer is:
(d) either plane or convex.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length $50 \mathrm{~cm}$.
(b) A concave lens of focal length $50 \mathrm{~cm}$.
(c) A convex lens of focal length $5 \mathrm{~cm}$.
(d) A concave lens of focal length $5 \mathrm{~cm}$.
To read small letters found in a dictionary, you would prefer to use a lens that magnifies the letters, making them appear larger and easier to read. Among the options given:
A convex lens converges light rays and can be used as a magnifying glass. The magnifying power of a lens is inversely proportional to its focal length, meaning a lens with a shorter focal length provides a higher magnification.
A concave lens diverges light rays and is not typically used for magnification purposes.
Given these points, the best option for magnifying and reading small letters would be:
(c) A convex lens of focal length (5 \mathrm{~cm}).
This lens would provide a higher magnification compared to a convex lens with a longer focal length (50 cm), making it the most suitable choice for reading small letters.
We wish to obtain an erect image of an object, using a concave mirror of focal length $15 \mathrm{~cm}$. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
To obtain an erect image using a concave mirror, the object must be placed between the focal point (F) and the mirror.
Given:
Focal length of the concave mirror, $f = -15 , \text{cm}$
To Find:
The range of distance of the object from the mirror.
The nature of the image.
Is the image larger or smaller than the object?
Ray diagram for image formation.
1. Range of Distance of the Object from the Mirror:
For a concave mirror, to get an erect image, the object has to be within the focal length. Therefore, the range of distance of the object from the mirror is from the mirror surface (0 cm) up to the focal point, which is 15 cm away from the mirror. So, the range is 0 cm < object distance < 15 cm.
2. Nature of the Image:
The image formed is virtual because it forms on the same side as the object, where light does not actually converge.
The image is erect, meaning it is upright as perceived by the viewer.
3. Is the Image Larger or Smaller than the Object?
The image is larger than the object. This is because, when the object is placed within the focal length of a concave mirror, the mirror creates a magnified, virtual, and erect image.
4. Ray Diagram:
To illustrate the image formation with a ray diagram:
A ray parallel to the principal axis reflects back through the focal point (F).
A ray passing through the focal point becomes parallel to the principal axis after reflection.
A ray heading towards the center of curvature reflects back on itself.
Since we're discussing a virtual image formation (the object is within the focal length), the reflected rays diverge. To find the image, we extend the reflected rays backward. They appear to meet at a point behind the mirror, forming a virtual, erect, and enlarged image.
AB is the object placed between the focus (F) and the mirror.
A'B' is the virtual, erect, and magnified image formed.
The rays diverge after reflection, but when extended backward, they seem to converge at A'B', forming the image.
This ray diagram represents the conditions under which an erect, magnified, and virtual image can be formed using a concave mirror.
Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
The type of mirrors used in various situations can be identified based on the kind of image they need to produce and the functionality required in each application.
(a) Headlights of a car: The type of mirror used in the headlights of a car is a concave mirror.
Reason: Concave mirrors can converge light beams to a focal point. In car headlights, the bulb is placed at the focal point of the concave mirror, causing the light rays to project outward in a parallel beam. This creates a focused beam of light that illuminates the road ahead without causing dispersion of light, which is essential for driving in the dark.
(b) Side/rear-view mirror of a vehicle: The type of mirror used in the side or rear-view mirrors of a vehicle is a convex mirror.
Reason: Convex mirrors have the ability to diverge light rays, which allows them to create a wider field of view. In the case of vehicle side or rear-view mirrors, this characteristic is beneficial as it allows drivers to see a larger area behind and to the sides of the vehicle, helping to reduce blind spots and improve safety while driving.
(c) Solar furnace: The type of mirror used in a solar furnace is a concave mirror.
Reason: Concave mirrors can concentrate parallel rays of light (such as sunlight) onto a single focal point. In a solar furnace, concave mirrors are used to focus sunlight onto a particular point where the furnace is located. This concentration of solar energy at the focal point produces very high temperatures, which can be used to heat materials or to generate electricity. This functionality exploits the converging properties of concave mirrors to maximize solar energy absorption.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations. 120 words
Yes, a convex lens covered with black paper on one-half will still produce a complete image of the object, although the image may be less bright. This phenomenon can be attributed to the principle of refraction and the ability of the lens to focus light. When light from an object passes through the uncovered half of the convex lens, it refracts (bends) towards the focal point on the other side of the lens. Due to the lens's shape, it can focus the light from the object into a complete image on a screen placed at the focal distance, despite half of the lens being covered.
Experimental Verification: Place an object in front of a convex lens and observe the image formed on a screen placed on the other side of the lens at the focal length. Now, cover one-half of the lens with black paper and adjust the screen until a clear image is obtained. You will observe that the image still forms, though it may appear dimmer due to the reduced light passing through the lens. This demonstrates that even with half of the lens covered, the lens can focus the light rays from the object to form a complete image.
Observations: The crucial point is that the completeness of the image formation by a lens is determined by its ability to bend light rays to a focal point, which is preserved even if part of the lens is obscured. The brightness or intensity of the image, however, depends on the amount of light passing through the lens, which in this case is reduced.
An object $5 \mathrm{~cm}$ in length is held $25 \mathrm{~cm}$ away from a converging lens of focal length $10 \mathrm{~cm}$. Draw the ray diagram and find the position, size and the nature of the image formed.
To solve this problem and draw the ray diagram, we'll follow the step-by-step process typical for lens ray diagrams, and we'll also calculate the position, size, and nature of the image formed by the converging lens. I'll start with the calculations using the lens formula and magnification formula.
Step 1: Use the Lens Formula
The lens formula is given by: $$ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $$ where (f) is the focal length of the lens, (v) is the image distance (from the lens to the image), and (u) is the object distance (from the lens to the object). The distances are positive if they are on the same side as the object, and negative if they are on the opposite side. For a converging lens, the focal length (f) is positive.
Given:
Focal length (f = 10, \text{cm})
Object distance (u = -25, \text{cm}) (negative since we're considering the direction towards the lens from the object to be negative)
Plug the values into the lens formula: $$ \frac{1}{10} = \frac{1}{v} - \frac{1}{-25} $$
Let's solve this equation for (v).
Step 2: Calculate the Image Distance (v)
Rearrange the lens equation to solve for (v): $$ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} $$ $$ \frac{1}{v} = \frac{1}{10} - \frac{1}{25} $$ Substitute the values and we get: $$ \frac{1}{v} = \frac{5 - 2}{50} = \frac{3}{50} $$ Hence, (v = \frac{50}{3} \approx 16.67, \text{cm})
Step 3: Calculation of the Image Size Using Magnification Formula
The magnification (M) is given by the ratio of the image height (h_i) to the object height (h_o), and also by the ratio of image distance (v) to object distance (u): $$ M = \frac{h_i}{h_o} = \frac{v}{u} $$ Given the object height (h_o = 5, \text{cm}), we can find the image height: $$ M = \frac{16.67}{-25} = -\frac{2}{3} $$ Thus, (h_i = 1.67, \text{cm}), and the image is inverted because the image height came out as negative.
Conclusion:
Image Position: (16.67, \text{cm}) from the lens, on the opposite side of the object.
Image Size: (1.67, \text{cm}) (inverted)
Nature of the Image: Real and inverted (since (v) is positive and (h_i) is negative).
Ray Diagram
For the ray diagram:
A ray parallel to the principal axis passes through the lens and emerges converging toward the focal point on the opposite side.
A ray passing through the center of the lens will travel straight through without bending.
A ray heading towards the focal point on the object's side of the lens emerges parallel to the principal axis on the other side.
Due to my current capabilities, I cannot directly draw or display images. To visualize this, you may apply the above steps on paper, noting that the image will form on the opposite side of the lens from the object, and it will be smaller and inverted compared to the object.
A concave lens of focal length $15 \mathrm{~cm}$ forms an image $10 \mathrm{~cm}$ from the lens. How far is the object placed from the lens? Draw the ray diagram.
To find the object distance ((u)) for a concave lens given the image distance ((v)) and the focal length ((f)), we can use the lens formula, which is given by:
$$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$
Given values:
Focal length ((f)) = -15 cm (Negative sign because concave lenses have negative focal lengths)
Image distance ((v)) = -10 cm (Negative sign because the image formed by a concave lens is virtual and on the same side as the object)
Now, let's solve for (u) using the lens formula:
Rearrange the lens formula to solve for (u):
$$ \frac{1}{u} = \frac{1}{f} - \frac{1}{v} $$
Substitute the given values:
$$ \frac{1}{u} = \frac{1}{-15} - \frac{1}{-10} $$
$$ \frac{1}{u} = -\frac{1}{15} + \frac{1}{10} $$
Calculate the value:
$$ \frac{1}{u} = -\frac{2}{30} + \frac{3}{30} $$
$$ \frac{1}{u} = \frac{1}{30} $$
Therefore, the object distance ((u)) is:
$$ u = 30 \mathrm{~cm} $$
Drawing the Ray Diagram for a Concave Lens:
Step 1: Draw the principal axis (a straight horizontal line).
Step 2: Mark the optical center of the lens (point O) on the principal axis.
Step 3: Position the focal points (F1 and F2) on both sides of the lens at the given focal length of 15 cm away from the lens. For a concave lens, the focal points are virtual and denoted on the same side as the object.
Step 4: Locate the object (a vertical arrow) on the principal axis, 30 cm away from the optical center (on the left side of the lens for a concave lens).
Step 5: Draw the first ray parallel to the principal axis from the top of the object, which after passing through the lens, diverges as if it is coming from the focal point (F1) on the same side as the object.
Step 6: Draw the second ray passing through the optical center of the lens without any deviation.
Step 7: The point where the backward extensions of these rays meet (above the principal axis on the same side as the object) marks the position of the virtual image formed by the concave lens. The image will be upright and smaller than the object.
Unfortunately, I cannot directly draw diagrams. For visual reference, it is best to consult a physics textbook or use online resources that specialize in optical diagrams.
An object is placed at a distance of $10 \mathrm{~cm}$ from a convex mirror of focal length $15 \mathrm{~cm}$. Find the position and nature of the image.
To find the position of the image formed by a convex mirror and the nature of the image, we can use the mirror equation: $$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$
Where:
(f) is the focal length of the mirror.
(v) is the image distance from the mirror (the value we're trying to find).
(u) is the object distance from the mirror.
Given that the object is placed (10 , \mathrm{cm}) from the mirror ((u = -10 , \mathrm{cm})), and the focal length of the convex mirror is (f = 15 , \mathrm{cm}) (for convex mirrors, the focal length is considered positive), we can substitute these values into the mirror equation.
Note: In mirror equations, distances measured against the direction of incident light are taken as negative. Thus, (u = -10 , \mathrm{cm}), but since it's a convex mirror, the focal length (f) remains positive.
Let's substitute the given values into the mirror equation and solve for (v). Remember to use the correct signs for the given values.
Rearranging the formula to solve for (v) gives us: $$ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} $$
Substituting the given values: $$ \frac{1}{v} = \frac{1}{15} - \frac{1}{-10} $$ The calculation gives us: $$ \frac{1}{v} = \frac{1}{6} $$
Therefore, the image distance (v) is the reciprocal of (1/6), which is (6 , \mathrm{cm}).
This means the image is formed (6 , \mathrm{cm}) away from the mirror on the same side as the object (since it's a convex mirror, the image formed is always virtual, and the distance is considered positive when measured in the same direction as the light reflected by the mirror).
Nature of the Image:
The image formed by a convex mirror is always virtual (as it cannot be projected onto a screen), upright (not inverted), and reduced in size compared to the object.
The magnification produced by a plane mirror is +1 . What does this mean?
The magnification produced by a mirror or lens tells us how much larger or smaller the image is compared to the object. A magnification of +1, as produced by a plane mirror, means two things:
Size: The image is the same size as the object (magnification of 1).
Orientation: The positive sign indicates that the image is upright relative to the object.
So, when a plane mirror produces a magnification of +1, it means the image you see in the mirror is upright and exactly the same size as the actual object. Additionally, the image formed by a plane mirror is virtual, meaning it appears to be behind the mirror and cannot be projected onto a screen.
An object $5.0 \mathrm{~cm}$ in length is placed at a distance of $20 \mathrm{~cm}$ in front of a convex mirror of radius of curvature $30 \mathrm{~cm}$. Find the position of the image, its nature and size.
To solve this problem, we need to use the mirror formula and the magnification formula. The mirror formula is given by:
$$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$
where \(f\) is the focal length of the mirror, \(v\) is the image distance (distance of the image from the mirror), and \(u\) is the object distance (distance of the object from the mirror). For a convex mirror, the focal length \(f\) is positive and is given by \(f=\frac{R}{2}\), where \(R\) is the radius of curvature of the mirror.
The magnification (m) is given by:
$$ m = \frac{h'}{h} = \frac{v}{u} $$
where (h') is the height of the image, (h) is the height of the object, (v) is the image distance, and (u) is the object distance. For this problem, we know:
- The object height (h = 5.0 , \text{cm})
- The object distance (u = -20 , \text{cm}) (The negative sign is because, by convention, distances measured against the direction of incident light are taken as negative)
- The radius of curvature (R = 30 , \text{cm}) (For convex mirrors, (R) is positive)
First, let's calculate the focal length (f) of the mirror:
$$ f = \frac{R}{2} = \frac{30}{2} = 15 , \text{cm} \text{ (positive for a convex mirror)} $$
Now, we can use the mirror formula to find the image distance (v). Substituting the known values into the mirror formula:
$$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$
$$ \frac{1}{15} = \frac{1}{v} + \frac{1}{-20} $$
Solving for (v):
$$ \frac{1}{v} = \frac{1}{15} - \frac{1}{-20} = \frac{1}{15} + \frac{1}{20} $$
Now, let's calculate the value of (v), and then we will discuss the nature and size of the image. Solving the equation, we find the image distance is:
$$ v = \frac{60}{7} \approx 8.57 , \text{cm} $$
Since \(v\) is positive, the image is formed on the same side as the outgoing light (i.e., behind the mirror for a convex mirror).
Nature of the Image:
Convex mirrors always form virtual, upright, and diminished images irrespective of the position of the object in front of them. Therefore, the image formed is virtual, upright, and diminished.
Size of the Image:
To find the size (height) of the image, we use the magnification formula:
$$ m = \frac{h'}{h} = \frac{v}{u} $$
Given \(h = 5.0 , \text{cm}) and the calculated and given values of \(v = 8.57 , \text{cm}) and \(u = -20 , \text{cm}), we can find (h'):
$$ m = \frac{8.57}{-20} \approx -0.4285 $$
The negative sign of magnification in the mirror indicates the image is formed in the opposite orientation to that of the object. However, for convex mirrors (considering real-life applications), we interpret magnification as positive because they always form virtual and upright images. So, we'll consider magnification's absolute value for size calculation:
$$ |h'| = |m| \times h = 0.4285 \times 5.0 \approx 2.1425 , \text{cm} $$
Thus, the size of the image is approximately \(2.14 , \text{cm}\), and it is diminished compared to the object.
Summary:
- Position of the image: Approximately \(8.57 , \text{cm}\) behind the mirror.
- Nature of the image: Virtual, upright, and diminished.
- Size of the image: Approximately \(2.14 , \text{cm}\).
An object of size $7.0 \mathrm{~cm}$ is placed at $27 \mathrm{~cm}$ in front of a concave mirror of focal length $18 \mathrm{~cm}$. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
To solve this problem, we will use the mirror formula and the magnification formula for a concave mirror. The mirror formula is given by:
$$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$
Where:
(f) is the focal length of the mirror,
(v) is the image distance from the mirror (positive if the image is on the same side as the object for concave mirrors, which means it's real; negative if on the opposite side, which implies it's virtual),
(u) is the object distance from the mirror (always taken as negative in mirror equations according to the sign convention).
The magnification ((m)) of the mirror is given by:
$$ m = \frac{h_i}{h_o} = \frac{-v}{u} $$
Where:
(h_o) is the height of the object,
(h_i) is the height of the image (positive if the image is upright and negative if inverted).
Given:
(f = -18 \mathrm{~cm}) (focal length is negative for concave mirrors),
(u = -27 \mathrm{~cm}) (object distance is always negative in mirror equation),
(h_o = 7.0 \mathrm{~cm}).
Let's first find the image distance ((v)) using the mirror formula:
$$ \frac{1}{-18} = \frac{1}{v} + \frac{1}{-27} $$
Solving this equation, we will find the value of (v).
Next, we use the magnification formula to find the size of the image ((h_i)):
$$ m = \frac{h_i}{7.0} = \frac{-v}{-27} $$
Solving for (h_i) will give us the size of the image. The nature of the image (whether it's real or virtual, upright or inverted) can be determined by the sign of (v) and (h_i).
Let's calculate (v) first. The calculation gives us the image distance (v = -54 , \mathrm{cm}). This indicates that the image is formed at (54 , \mathrm{cm}) on the same side of the mirror as the object, which means the image is real and inverted since (v) is negative according to the sign convention used in optics for concave mirrors.
Now, let's calculate the size of the image ((h_i)) using the magnification formula:
$$ m = \frac{h_i}{7.0} = \frac{-v}{-27} $$
Substituting (v = -54) into the equation, we can solve for (h_i). The magnification ((m)) is (2), which means the image is twice the size of the object. Using the magnification formula (m = \frac{h_i}{h_o}), we can find the size of the image ((h_i)):
Given (m = 2) and (h_o = 7.0 , \mathrm{cm}), we have:
$$ 2 = \frac{h_i}{7.0} $$
Solving for (h_i):
$$ h_i = 2 \times 7.0 = 14.0 , \mathrm{cm} $$
Therefore, the size of the image is (14.0 , \mathrm{cm}), and since (m) is positive, the image is real and inverted. Additionally, the fact that (h_i > h_o) and (v < 0) confirms that the image is both magnified and inverted.
In summary:
The screen should be placed 54 cm from the mirror to obtain a sharp focused image.
The size of the image is 14.0 cm and it is inverted and magnified.
Find the focal length of a lens of power $-2.0 \mathrm{D}$. What type of lens is this?
The power of a lens ( P ) is related to its focal length ( f ) by the formula:
$$ P = \frac{1}{f} $$
where ( P ) is measured in diopters ((\mathrm{D})) and ( f ) is measured in meters ((m)).
Given the power ( P = -2.0 \mathrm{D} ), we can find the focal length ( f ) by rearranging the formula:
$$ f = \frac{1}{P} $$
Substituting the given power:
$$ f = \frac{1}{-2.0} $$
$$ f = -0.5 , \mathrm{m} $$
The focal length of the lens is (-0.5 , \mathrm{m}), which is a negative value.
A negative focal length indicates a diverging lens, which means this lens is a concave lens.
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
The power of a lens ((P)) in diopters (D) is the inverse of its focal length ((f)) in meters:
$$ P = \frac{1}{f} $$
Given that the power of the prescribed lens is (+1.5\ \text{D}), we can find the focal length ((f)) by rearranging the formula:
$$ f = \frac{1}{P} $$
Substituting the given power:
$$ f = \frac{1}{+1.5} = \frac{2}{3} \text{ meters} $$
So, the focal length of the lens is (\frac{2}{3}\ \text{m}) or (0.67\ \text{m}).
The sign of the power tells us whether the lens is converging or diverging. A positive power indicates a converging lens, which means the prescribed lens is a converging lens.
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Ask Chatterbot AIExtra Questions and Answers - Light – Reflection and Refraction | NCERT | Science | Class 10
Assume the distance between the sun and the mirror is infinity, and the sun is much above the principal axis. Where will the image of the sun be formed?
A. at $f$
B. at $R$
C. in the focal plane but below the principal axis
D. in the focal plane but above the principal axis
The correct answer is option C: in the focal plane but below the principal axis.
Given that the Sun is at infinity, the incoming sun rays on the mirror are parallel to the principal axis. By the properties of mirrors, parallel rays converge at the focal point. Since the sun's position is specified as "much above the principal axis," the parallel rays are directed downward, causing the image to form in the focal plane but below the principal axis.
What is meant by the 'angle of incidence' and the 'angle of refraction' for a ray of light?
The angle of incidence is defined as the angle formed between the incident ray and the normal (a line perpendicular to the surface) at the point where the light ray hits the surface. This angle is often denoted as $i$.
Similarly, the angle of refraction is the angle between the refracted ray and the normal at the point where the light ray enters the new medium. It is typically represented by $r$.
What happens when a ray of light passes through the optical centre of a lens?
A. No light ray can pass through the optical centre.
B. It passes through the lens undeviated.
C. It becomes parallel to the principal axis.
D. None of the above."
The correct option is B. When a ray of light passes through the optical center of a lens, it passes through the lens undeviated. This is due to the symmetry and the way lenses are designed, allowing light that enters through the optical center to not be bent by the lens materials. Thus, option B accurately describes this behavior.
Why do we use tungsten filament in bulbs or lamps?
Tungsten is the preferred choice for filaments in bulbs and lamps primarily due to its high melting point, which is around $3380^\circ \text{C}.$
This incredibly high melting point ensures that tungsten does not melt under the intense heat generated when the filament glows. Tungsten becomes incandescent, meaning it emits light, at about $2400 \text{ K},$ making it ideal for producing light without melting.
What is the nature of a lens having a power of $+0.5 \mathrm{D}$?
The nature of a lens is determined by its power value. In this case, the power of the lens is given as $+0.5 \mathrm{D}$.
Since the power is positive, it indicates that the focal length of the lens is positive. Lenses with a positive focal length are converging lenses. Therefore, the lens with a power of $+0.5 \mathrm{D}$ is a converging lens.
In a pinhole camera, the image of height $6 \mathrm{~cm}$ is formed at a distance of $10 \mathrm{~cm}$. If the object is situated at a distance of $5 \mathrm{~cm}$, the height of the object is:
A. $3 \mathrm{~cm}$
B. $6 \mathrm{~cm}$
C. $1.5 \mathrm{~cm}$
D. None of these
The correct option is A. $3 \mathrm{~cm}$.
In a pinhole camera, magnification ($m$) is given by the ratio of the image distance ($v$) to the object distance ($u$): $$ m = \frac{v}{u} = \frac{10 \text{ cm}}{5 \text{ cm}} = 2 $$
Since magnification is also the ratio of the image height to the object height: $$ m = \frac{\text{Height of image}}{\text{Height of object}} $$ Given that $m = 2$ and the height of the image is $6$ cm, we can find the height of the object: $$ \frac{6 \text{ cm}}{\text{Height of object}} = 2 \Rightarrow \text{Height of object} = \frac{6 \text{ cm}}{2} = 3 \text{ cm} $$
Thus, the height of the object is 3 cm.
Which of the following objects is NOT transparent?
A) A thick colored glass window
B) Air
C) Vacuum
D) Water
The correct option is A: A thick colored glass window.
Among the given options, a thick colored glass window is not transparent because it does not allow for clear visibility of objects on the other side; it partially blocks and absorbs light. This type of material is described as translucent rather than transparent. Thus, you cannot see through it clearly.
Explain why the planets do not twinkle?
The twinkling of stars is attributed to their being point sources of light combined with atmospheric refraction. Since stars are extremely distant, they appear as mere points in the sky. As the light from these stars passes through the Earth's atmosphere, it undergoes refraction at various layers, which differ in air density. The apparent increase or decrease in the star's brightness, caused by the varying paths of light reaching the observer, leads to the twinkling effect.
On the other hand, planets do not twinkle because they are closer to Earth and thus appear larger than stars. This size difference means planets can be thought of as having many point-like light sources grouped together. As these points of light pass through the atmosphere, the effects of refraction that cause twinkling are averaged out across these multiple sources. The contrasting dimming and brightening across different parts of a planet’s apparent disc average out, nullifying the twinkling effect. Thus, the overall appearance of a planet remains steady and without twinkle when viewed from Earth.
What is responsible for looming?
A) Total internal reflection of light
B) Dispersion of light
C) Coherence of light
D) Refraction of light.
The correct option is A. Total internal reflection of light
Looming is an optical phenomenon typically observed on a sea-shore during a winter evening. This phenomenon causes the image of a ship to appear as if it is floating in the air, where the sea meets the sky. The primary cause of this effect is total internal reflection of light, which manipulates the way the image is perceived.
When a concave mirror is used as a shaving mirror, where is the person's face in relation to the focus of the mirror?
A. At focus
B. Between the centre of curvature and focus
C. Between pole and focus
D. Beyond the centre of curvature
The correct answer is C. Between pole and focus
When using a concave mirror as a shaving mirror, it's important to position the face between the pole and the focus of the mirror. This placement ensures that the mirror provides an upright and magnified image, which is ideal for shaving or applying makeup. Concave mirrors used for this purpose typically have a large focal length to ensure that the face can comfortably be positioned between the focus and the pole, and a large aperture to allow viewing of the entire face clearly.
Why is the focal length of a plane mirror infinite?
A plane mirror has a flat or planar reflective surface, causing it to differ significantly from spherical mirrors which have curved surfaces. The definition of a sphere's curvature and radius is pivotal yet doesn't apply as straightforwardly here since a plane surface technically doesn't curve.
In optics, the focal length ( f ) is calculated as half the radius of curvature (( R )) of the mirror:
$$ f = \frac{R}{2} $$
Because a plane mirror does not curve, its radius of curvature is considered to be infinite:
$$ f = \frac{\text{infinite}}{2} = \text{infinite} $$
This implies that parallel rays reflecting off a plane mirror never actually converge or diverge (as they would with a curved mirror); they remain parallel even after reflection, suggesting an infinite focal length.
Further, from the mirror equation, where ( v ) is the image distance and ( u ) is the object distance:
$$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$
But for a plane mirror, the image distance ( v ) is equal and opposite to the object distance ( u ) ( ( v = -u ), when considering sign convention):
$$ \frac{1}{f} = \frac{1}{-u} + \frac{1}{u} = 0 $$
Hence, by solving,
$$ f = \frac{1}{0} = \text{infinite} $$
Fundamentally, the infinity focal length highlights that the image is formed at an infinite distance behind the mirror, aligning with the concept that images formed by plane mirrors are virtual and erect.
An object is placed at a distance of $25$ cm from the pole of a spherical mirror which forms a real, inverted image on the same side of the object at $31.5$ cm from the pole. Calculate the focal length of the mirror and find the nature of the mirror.
The problem provides the following details:
The object distance ($u$) is $25$ cm. Since we measure object distance as negative in mirror formula, $u = -25$ cm.
The image distance ($v$) is given as $31.5$ cm, and since the image forms on the same side as the object (being real and inverted), it is negative, giving us $v = -31.5$ cm.
We use the mirror equation to calculate the focal length ($f$):
$$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$
Plugging in the values:
$$ \frac{1}{f} = \frac{1}{-31.5} + \frac{1}{-25} $$
$$ \frac{1}{f} = -0.03175 - 0.04 = -0.07175 $$
Thus, $f = \frac{1}{-0.07175} \approx -13.93 $ cm.
Rounding off, we get $f \approx -14$ cm.
This negative focal length indicates that the mirror is concave.
What is $2~\mathrm{F}$ called in simple words?
In optics, the relationship between the radius of curvature $(R)$ and the focal length $(F)$ of a mirror is given by: $$ R = 2F $$ Here, if we consider $F = 1$, then $R$ would be $2 \times 1 = 2$. Therefore, $2F$ often signifies the radius of curvature of a mirror when $F$ is the focal length.
Understanding this in terms of geometry, the center of curvature of a curve (or mirror surface in optics) lies at a distance from the curve that equals the radius of curvature $(R)$, positioned along the curve's normal. The center of curvature serves as the center for the osculating circle of the curve.
Name the spherical mirror which (i) diverges and (ii) converges the beam of light incident on it. Justify your answer by drawing a ray diagram in each case.
(i) Diverges
A convex mirror diverges a beam of light falling on it. In other words, when light rays strike a convex mirror, they reflect outwards, causing the rays to spread apart.
Ray Diagram for Convex Mirror:
Light rays, originating from a point, strike the mirror.
After reflection, the rays appear to diverge from a point behind the mirror, suggesting a virtual, upright, and diminished image.
(ii) Converges
A concave mirror converges a beam of light falling on it. This means that the light rays coming towards a concave mirror are reflected such that they meet at a point (or appear to come from a point if extended backward).
Ray Diagram for Concave Mirror:
Light rays, approaching parallel to the principal axis, reflect and meet at the focal point of the mirror.
This behavior of concave mirrors enables them to form real, inverted images or virtual, upright images depending on the position of the object relative to the focal point.
Which mirror always forms a virtual image?
A. Concave
B. Convex
C. Both concave and convex
D. Neither concave nor convex
The correct answer is B. Convex
Convex mirrors diverge light rays that originate from a point source. Since these rays do not meet at any point in front of the mirror, it is necessary to extend them behind the mirror to determine their apparent meeting point. As the image location is behind the mirror, it cannot be projected onto a screen and hence is not a real image but a virtual image. Convex mirrors consistently form virtual images.
A double concave lens is placed in water of refractive index $\frac{4}{3^{3}}$. If the radii of curvature of the two surfaces of the lens are $10 \mathrm{~cm}$ and $30 \mathrm{~cm}$; and the refractive index of glass is $\frac{3}{2}$, find the focal length of the lens in water.
A) $60 \mathrm{~cm}$
B) $-60 \mathrm{~cm}$
C) $30 \mathrm{~cm}$
D) $-30 \mathrm{~cm}$
The correct answer is Option B: $-60 \mathrm{~cm}$.
To find the focal length of the lens in water, we use the lens maker's formula for a lens in a medium: $$ \frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) $$ where $\mu_2$ and $\mu_1$ are the refractive indices of the lens material and the surrounding medium respectively, and $R_1$ and $R_2$ are the radii of curvature of the two surfaces of the lens.
Given:
The refractive index of glass $\mu_2 = \frac{3}{2}$
The refractive index of water $\mu_1 = \frac{4}{3}$
The radii of curvature of the lens surfaces $R_1 = -10 \mathrm{~cm}$ (since it is double concave, the radius of the first surface facing the incoming light is negative) and $R_2 = 30 \mathrm{~cm}$
Substituting these values into the formula: $$ \begin{align*} \frac{1}{f} &= \left(\frac{3}{2} \times \frac{3}{4} - 1\right)\left(\frac{1}{-10} - \frac{1}{30}\right) \ &= \left(\frac{9}{8} - 1\right)\left(-\frac{1}{10} + \frac{1}{30}\right) \ &= \left(\frac{1}{8}\right)\left(-\frac{3}{30} + \frac{1}{30}\right) \ &= \left(\frac{1}{8}\right)\left(-\frac{2}{30}\right) \ &= -\frac{1}{120} \ f &= -60 \mathrm{~cm} \end{align*} $$
Thus, the focal length of the lens in water is $-60 \mathrm{~cm}$. This negative sign indicates that it is a diverging lens, which is consistent with the double concave shape of the lens.
What happens when a ray of light gets refracted through a concave lens passing through F1?
Solution:When a ray of light passes through the focal point $F_1$ of a concave lens, it will refract and diverge away from the principal axis. Hence, after refraction, the ray of light will bend farther away from the principal axis.
According to Snell's law, the value of the constant (refractive index) depends on:
A. The pair of media
B. The pair and order of media (where order implies from where to where)
C. The angle of incidence
D. The angle of refraction
The correct option is B: The pair and order of media (where order implies from where to where).
Snell's law, which governs the bending of light at the interface between two media, states that the refractive index depends not just on the pair of media involved, but also on the order in which the light passes through these media. This means that whether light travels from medium I to medium II or from medium II to medium I can influence the constant's value. Each distinct pair and order of transition will exhibit a unique refractive index. Thus, the value of the constant varies accordingly.
The angle at which sunlight rays hit the Earth's surface is known as
A. Critical Angle
B. Angle of Incidence
C. Ray angle
D. Angle of Reflection
The correct answer is B. Angle of Incidence.
The Angle of Incidence refers to the angle at which sunlight rays strike the Earth's surface. This angle is defined as the angle between the incoming light ray and the normal (perpendicular) to the surface at the point of contact.
Which of the following statement(s) is/are correct for reflection of light?
Statement 1: Angle of incidence is greater than the angle of reflection.
Statement 2: Angle of incidence is less than the angle of reflection.
Statement 3: Angle of incidence is equal to the angle of reflection.
Statement 4: The incident ray, the normal at the point of incidence, and the reflected ray all lie in the same plane.
A 1 and 4
B 2 and 4
C 3 and 4
D Only 3
The correct answer to the given question is C) 3 and 4.
According to the laws of reflection:
The angle of incidence is equal to the angle of reflection. This means that the angle formed between the incident ray and the normal to the surface at the point of incidence is exactly equal to the angle formed between the reflected ray and the same normal.
The incident ray, the normal at the point of incidence, and the reflected ray all lie in the same plane. This second law ensures that all these vectors are coplanar and thus, the reflection happens in a predictable and consistent manner.
Both Statement 3 and Statement 4 align perfectly with the established laws of reflection, making option C the correct choice.
Which of the following statement(s) is/are true regarding diffused reflection?
A. On a rough surface, light is reflected in different directions.
B. On a smooth surface, light is reflected in different directions.
C. Light doesn't reflect on a diffused surface.
D. Light is not completely absorbed on a diffused surface.
The correct answer is A: On a rough surface, light is reflected in different directions.
Explanation:
When light falls on a rough surface, the light rays are scattered in various directions due to the surfaces' irregularities. This phenomenon is known as diffused reflection. Contrarily, on a smooth surface, light tends to be reflected in a single, predictable direction, which is termed as regular reflection. Therefore, the statements B and C are incorrect. Statement D, regarding light not being completely absorbed on a diffused surface, although somewhat relevant, isn't fully addressing the concept of diffused reflection described in option A.
The refractive index of glass is $\frac{3}{2}$. The velocity of light in glass would be
A) $3 \times 10^{8} \mathrm{~ms}^{-1}$
B) $2 \times 10^{8} \mathrm{~ms}^{-1}$
C) $10^{8} \mathrm{~ms}^{-1}$
D) $1.33 \times 10^{8} \mathrm{~ms}^{-1}$
Solution:
The correct option is B) $2 \times 10^{8} \mathrm{~ms}^{-1}$.
The refractive index $n$ is defined as the ratio of the speed of light in vacuum to the speed of light in a medium. This relationship can be expressed as: $$ n = \frac{c}{v} $$ Hence, you can rearrange the formula to solve for velocity $v$ (the speed of light in the medium) as follows: $$ v = \frac{c}{n} $$ Given that the refractive index of glass $n = \frac{3}{2}$ and the speed of light in vacuum $c = 3 \times 10^{8} \mathrm{~ms}^{-1}$, substituting these values in the formula gives: $$ v = \frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{\frac{3}{2}} = 2 \times 10^{8} \mathrm{~ms}^{-1} $$ Thus, the velocity of light in glass is $2 \times 10^{8} \mathrm{~ms}^{-1}$.
An endoscope is employed by a physician to view the internal parts of a body organ. It is based on the principle of:
A) Refraction
B) Reflection
C) Total internal reflection
D) Dispersion
The correct answer is C) Total internal reflection.
An endoscope operates on the principle of total internal reflection. It consists typically of two or three main optical cables. Light travels down through one or two of these cables into the patient's body, and another cable carries the reflected light (the image from inside the body) back up to the physician's eyepiece. This mechanism allows the physician to view the internal organs of the body clearly through the endoscope.
What is the magnification produced by a mirror when the image formed is half the size of the object?
A) $\mathrm{m}=2$
B) $\mathrm{m}=0.5$
C) $\mathrm{m}=4$
D) $\mathrm{m}=0.25$
The correct answer is: B) $ \mathrm{m} = 0.5 $
The magnification ($m$) can be determined using the formula: $$ m = \frac{\text{Height of the image}}{\text{Height of the object}} $$
Assuming the height of the object is $x$, then the height of the image is half of that, which is $\frac{1}{2}x$ or $0.5x$. Substituting these values into the formula gives: $$ m = \frac{0.5x}{x} = 0.5 $$ This calculation shows the image size is half that of the object, corresponding to a magnification of 0.5.
Define red shift and blue shift in light.
Red shift refers to the phenomenon in physics where light emitted from an object moving away from the observer undergoes an increase in wavelength. This shift toward the longer wavelengths manifests as a change in color towards the red part of the spectrum, hence the term 'red shift'. This effect is a result of the Doppler effect and is notably observed in the light from distant galaxies. The presence of red shift in these galaxies indicates that the universe is expanding, thereby supporting the Big Bang theory.
Blue shift, on the other hand, occurs when an object emitting light is moving towards the observer. This results in a decrease in the wavelength and an increase in the frequency of the light waves. In visible light, this effect shifts the color from the red end of the spectrum towards the blue end. Blue shift is essentially the opposite of red shift.
What happens when a virtual object is used as an image in a convex mirror?
When a virtual object is used as the object for a convex mirror, the resulting image formed by the mirror will be real. This demonstrates that using virtual objects can indeed lead to the formation of real images with convex mirrors.
Blinking our eyes when light suddenly falls on our eye is a reflex action.
A. Reflex action
B. Voluntary action
The correct option is A. Reflex action
Blinking of the eyes in response to sudden light exposure is involuntary and spontaneous, categorized as a reflex action. This occurs as the eyes instinctively react to changes in light intensity to protect the retina, controlled by neural mechanisms outside of conscious control. This type of action is termed a cerebral reflex, governed by the cerebrum.
A circular object, when kept between a small source of light and a screen, casts a sharp shadow on the screen.
This experiment proves:
A. Rectilinear propagation of light
B. Laws of reflection
C. Wave nature of light
D. Laws of refraction
The correct answer is A. Rectilinear propagation of light.
When a circular object is placed between a small source of light and a screen, casting a sharp shadow, it demonstrates the rectilinear propagation of light. This principle means that light travels in straight lines. The formation of a sharp shadow occurs because the light rays from the source do not bend around the object; rather, they travel directly to the screen. If light did not travel in straight lines, these rays would curve around the object, filling the shadow region with light to an extent. Thus, the presence of a distinct, sharp shadow confirms that light travels in straight paths.
When light rays travel obliquely from a rarer medium to a denser medium, they slow down and bend towards the normal.
A) True
B) False
The correct answer is A) True.
When a light ray travels obliquely from a less dense (rarer) medium to a more dense (denser) medium, its speed decreases. This phenomenon is due to the fact that light travels at different speeds in different media. As a result of the slower speed in the denser medium, according to Fermat's Principle, the path of the light bends towards the normal. This bending of light when it passes from one medium to another is known as refraction.
Rainbow is always formed in a direction opposite to that of the sun.
A) True
B) False
The correct option is A) True
A rainbow is created when light undergoes total internal reflection inside a raindrop, causing the light to reflect back towards its original source. Consequently, rainbows are visible only when the observer is facing away from the sun, confirming that rainbows always appear in the direction opposite to the sun.
Which part of the solar cooker is responsible for the greenhouse effect?
A Coating with black color inside the box
B Mirror
C Glass sheet
D Outer cover of the solar cooker
The correct answer is C: Glass sheet.
The glass sheet in a solar cooker plays a crucial role in creating the greenhouse effect. It allows the majority of visible and ultraviolet solar radiation to pass through while absorbing most of the infrared radiation. This property means that sunlight can enter and directly heat the pot and the air inside the cooker. However, the thermal energy re-radiated by these heated objects cannot easily escape because the glass does not readily transmit infrared radiation back out. Hence, the glass traps the heat inside the cooker, mirroring the greenhouse effect. Thus, creating a warmer environment inside the solar cooker which helps in cooking the food efficiently.
No matter how far you stand from a mirror, your image appears erect. Which mirror are you most likely standing in front of?
A. Plane mirror
B. Concave mirror
C. Convex mirror
D. Either concave or convex mirror.
The correct answers are:
A. Plane mirror
C. Convex mirror
Both plane mirrors and convex mirrors consistently produce virtual and erect images regardless of the distance from which the observer views. Thus, the type of mirror in which your image appears erect, no matter the distance, is most likely either a plane or a convex mirror.
"What will happen if a ray of light strikes two mediums or surfaces?"
Solution
When a ray of light strikes two different media or surfaces at an angle, it undergoes refraction. This process causes the light ray to bend and alter its path. This bending of light leads to an optical illusion where objects, such as the bottom of a water tank or pond, appear to be located at positions other than where they actually are. This phenomenon occurs because the light rays seem to emanate from a point different from their actual source.
What happens to a white light when it passes through a prism?
When white light passes through a prism, it undergoes a process called dispersion. Dispersion occurs because light is composed of a spectrum of colors, each with a distinct wavelength. As each color passes through the prism, it bends at a slightly different angle, depending on its wavelength.
The result is that the components of white light — which include the colors red, orange, yellow, green, blue, indigo, and violet — separate into a visible spectrum often referred to as a rainbow. This occurs because the angle of refraction (the angle at which light bends) varies for different colors; red light bends the least while violet bends the most.
Hence, the passage of white light through a prism beautifully demonstrates that what we perceive as "white" light is actually a mixture of multiple colors. This effect is purely due to the physical properties of the prism, which acts to separate the colors that are already present in the white light due to their differing wavelengths.
"What is meant by dispersion of light? What is the cause of it?"
Dispersion of light is the process where a beam of white light is separated into its seven constituent colors upon passing through a transparent medium like a prism. The fundamental cause of dispersion lies in the fact that the refractive index of a material changes with the wavelength (and frequency) of light. This variability leads to different colors of light bending by varying amounts when they pass through a medium, thereby causing the light to split into a spectrum.
A coin is kept at the bottom of an empty beaker. Water having refractive index of $\frac{4}{3}$ is poured into the beaker up to a height of $16$ cm. Find the apparent position of the coin.
A) $10$ cm
B) $12$ cm
C) $7.5$ cm
D) $2.5$ cm
Solution
The correct answer is Option B: $12$ cm
Given:
The refractive index of water, $n = \frac{4}{3}$
Real depth, $d = 16$ cm
Relation between real and apparent depth due to refraction:$$ n = \frac{\text{Real Depth}}{\text{Apparent Depth}} $$ Rearranging to find the apparent depth, we have: $$ \text{Apparent Depth} = \frac{\text{Real Depth}}{n} $$
Substituting the given values: $$ \text{Apparent Depth} = \frac{16, \text{cm}}{\frac{4}{3}} = 12, \text{cm} $$
Thus, the apparent position of the coin is 12 cm below the surface.
Object is placed on the principal axis of a concave mirror. Its real image is formed at a point which is $10 \mathrm{~cm}$ more distant from the mirror than the object. If the focal length of the mirror is $12 \mathrm{~cm}$, the magnification of the image is:
A. 2
B. 1.5
C. 1
D. 2.5
Solution The correct choice is Option B: 1.5
Given values:
The focal length of the concave mirror, $f = -12 \text{ cm}$ (negative since it is a concave mirror).
Object distance, denoted by $u$ (note: since the object is in front of the mirror, $u$ is negative).
Image distance $v$, which is $10 \text{ cm}$ more than the object distance, hence $v = u + 10$.
Since we know the mirror formula: $$ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} $$ where:
$f$ is the focal length,
$u$ is the object distance, and
$v$ is the image distance.
Substituting $v = u + 10$: $$ \frac{1}{f} = \frac{1}{u} + \frac{1}{u + 10} $$ Simplifying with $f = -12$: $$ \frac{1}{-12} = \frac{1}{u} + \frac{1}{u + 10} $$
Rearranging terms and solving the quadratic equation: $$ \frac{1}{u} + \frac{1}{u+10} = \frac{1}{-12} $$ becomes: $$ u^2 - 14u - 120 = 0 $$ Factoring yields solutions: $$ (u - 20)(u + 6) = 0 $$ This gives $u = 20$ or $u = -6$. The convention for mirrors is that $u$ should be negative, considering the real object position, hence: $$ u = -20 \text{ cm} $$ Since $v = u + 10$, we find: $$ v = -20 + 10 = -30 \text{ cm} $$
The magnification (m) formula for mirrors is: $$ m = -\frac{v}{u} $$ Substituting the values of $v$ and $u$: $$ m = -\frac{-30}{-20} = -1.5 $$
Thus, the magnitude of magnification is: $$ |m| = 1.5 $$
This confirms that the correct answer is Option B: 1.5.
The speed of light in glass is $1.5 \times 10^{8} \mathrm{~m/s}$. Then the refractive index of glass is
A) 4 B) 3 C) 1 D) 2
The refractive index of a medium like glass is determined using the formula: $$ n = \frac{c}{v} $$ where:
$n$ is the refractive index,
$c$ is the speed of light in vacuum (approximately $3.0 \times 10^{8} , \text{m/s}$),
$v$ is the speed of light in the medium (for glass, it's given as $1.5 \times 10^{8} , \text{m/s}$).
Plugging in the known values: $$ n = \frac{3.0 \times 10^{8} , \text{m/s}}{1.5 \times 10^{8} , \text{m/s}} $$ $$ n = 2 $$
Therefore, the refractive index of glass is 2, making option D) 2 the correct answer.
Find the time taken by light to cover a distance of $9, \mathrm{mm}$ in water? Take $\mu_{\text{water}} = \frac{4}{3}$.
A) $0.04, \mathrm{ns}$
B) $0.4, \mathrm{ns}$
C) $4, \mathrm{ns}$
D) $400, \mathrm{ns}$
The correct answer is A) $0.04, \mathrm{ns}$.
First, we calculate the speed of light in water, $v$, using Snell's law, where the refractive index $\mu$ for water is given as $\frac{4}{3}$. The speed of light in air (vacuum) is approximately $3 \times 10^{8} \mathrm{~m/s}$. For air, $\mu_{\text{air}}$ is nearly 1.
Using the relationship: $$ \mu_{\text{air}} v_{\text{air}} = \mu_{\text{water}} v_{\text{water}} $$ Substitute the values: $$ 1 \times 3 \times 10^{8} = \frac{4}{3} \times v_{\text{water}} $$ Solving for $v_{\text{water}}$: $$ v_{\text{water}} = \frac{3 \times 10^{8} \times 3}{4} = \frac{9}{4} \times 10^{8} \mathrm{~m/s} $$
To find the time $t$ taken by light to cover a distance of $9, \mathrm{mm}$ (or $9 \times 10^{-3} \mathrm{~m}$) in water, use the formula: $$ t = \frac{d}{v} $$ Substitute the distance $d$ and the speed $v_{\text{water}}$: $$ t = \frac{9 \times 10^{-3} \mathrm{~m}}{\frac{9}{4} \times 10^{8} \mathrm{~m/s}} $$ Calculate $t$: $$ t = \frac{9 \times 10^{-3}}{\frac{9}{4} \times 10^{8}} = 4 \times 10^{-11} \mathrm{~s} $$ Convert seconds to nanoseconds (1 s = $10^9$ ns): $$ t = 0.04 \mathrm{~ns} $$
Thus, the time taken by light to travel $9, \mathrm{mm}$ in water is 0.04 nanoseconds.
Which of the following statements is correct regarding the propagation of light of different colors of white light in air?
A) Red light moves fastest.
B) Violet light bends the most.
C) Both (1) and (2).
D) All the colors of white light move with the same speed.
The correct answer is C) Both (1) and (2).
In air, different colors of light have slightly different speeds due to their wavelengths, which affects their refraction. Although the speed differences are minute because air is almost a vacuum, red light tends to travel slightly faster than other colors and undergoes less bending. In contrast, violet light, having a shorter wavelength, travels comparatively slower and is bent more than other colors when transitioning between different media. This bending is critical in applications like prisms and rainbows, where light is dispersed into its constituent colors.
Which of the following has the most abrasive or irregular surface?
A) Mirror
B) Stainless Steel
C) Sandpaper
D) Surface of an ice cube
The correct answer is C) Sandpaper.
Sandpaper is specifically designed with a highly abrasive surface to smooth and finish surfaces, primarily wood. Its rough surface is intended for sanding down materials, which is significantly more abrasive compared to the other options listed.
During a winter evening, the image of a ship is seen formed in the air in the sea-sky. This phenomenon is due to total internal reflection.
A) True
B) False
The correct answer is A) True.
This phenomenon is indeed due to total internal reflection. During a cold winter evening, the seawater close to the seabed becomes extremely cold, causing the air layer in contact with it to also become colder and denser. As altitude increases, the air becomes progressively warmer and less dense. Light rays from a ship, which may not be directly visible, travel upwards from this colder, denser air into the warmer, rarer air layers. When these light rays hit the boundary between the denser and rarer air at a sufficiently shallow angle, they undergo total internal reflection. As a result, the rays are reflected downward, and an observer on the seashore can see them. This creates a virtual image of the ship appearing to hang in the sky.
A concave lens has a focal length of $15 \mathrm{~cm}$. At what distance should the object be from the lens placed so that it forms an image $10 \mathrm{~cm}$ from the lens?
A) $30 \mathrm{~cm}$
B) $60 \mathrm{~cm}$
C) $90 \mathrm{~cm}$
D) $100 \mathrm{~cm}$
Solution:
The correct answer is Option A: $30 \mathrm{~cm}$.
For a concave lens, the focal length is negative, and it forms a virtual and erect image on the same side as the object. Given:
Image distance, $v = -10 \mathrm{~cm}$
Focal length, $f = -15 \mathrm{~cm}$
We need to find the object distance, $u$.
We use the lens formula: $$ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} $$ Plugging the given values into the formula: $$ \frac{1}{-10} - \frac{1}{u} = \frac{1}{-15} $$ Solving for $u$, we rearrange the equation: $$ -\frac{1}{u} = \frac{1}{-15} - \frac{1}{-10} = \frac{-2 + 3}{30} = \frac{1}{30} $$ Therefore: $$ \frac{1}{u} = -\frac{1}{30} \Rightarrow u = -30 \mathrm{~cm} $$
The negative sign indicates that the object is on the same side as the usual direction of light travel, which is typical for a concave lens.
Thus, the object should be placed $30 \mathrm{~cm}$ from the lens.
"You are given a transparent glass sheet. Suggest any two ways to make it translucent without breaking it."
Solution
Transparent objects permit light to completely pass through them, whereas translucent objects allow light to partially pass through.
Here are two ways to convert a transparent glass sheet into a translucent one without breaking it:
Coat its one side with grease or oil. This method disrupts the clear transmission of light due to the opacity introduced by the layer of grease or oil.
Rub the glass sheet with an abrasive material. Using an abrasive material to scratch the surface of the glass will create micro-abrasions that scatter the light, reducing the glass's transparency and increasing its translucency.
Match the following: Column A Column B (a) White light (i) Convex mirror (b) Refraction (ii) Concave mirror (c) Virtual images (iii) Refraction (d) Real images (iv) Spectrum (e) Prism (v) Ray of light from glass to air
Ans. (a) - (iv), (b) - (v), (c) - (i), (d) - (ii), (e) - (iii)
The correct matches for the given terms in Column A to their associated phenomena in Column B are as follows:
(a) White light corresponds to (iv) Spectrum. This is because when white light passes through a prism, it disperses into a spectrum of colors.
(b) Refraction matches with (v) Ray of light from glass to air. This scenario effectively exemplifies refraction, where the bending of light occurs as it passes from one medium to another of different optical density.
(c) Virtual images are typically associated with (i) Convex mirror. A convex mirror always forms virtual images because the reflected rays diverge and only appear to come from a focal point behind the mirror.
(d) Real images are accurately linked to (ii) Concave mirror. A concave mirror can form real images when the object is placed outside the focal distance of the mirror, causing converging reflected rays that meet.
(e) Prism has a direct relation with (iii) Refraction. A prism utilizes refraction to disperse light into its constituent spectral colors, demonstrating the effect when light passes through it.
An object is placed $15$ cm in front of a convex lens of focal length $10$ cm. Find the nature and position of the image formed. Where should a concave mirror of radius of curvature $20$ cm be placed so that the final image is formed at the position of the object itself?
Solution:
Step 1: Finding the image formed by the convex lensGiven:
Object distance, $ u = -15 \text{ cm} $
Focal length of the convex lens, $ f = 10 \text{ cm} $
Using the lens formula: $$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$ where,
$ f $ is the focal length of the lens,
$ u $ is the object distance,
$ v $ is the image distance.
Substitute the known values: $$ \frac{1}{v} = \frac{1}{10} - \frac{1}{15} $$ Finding the common denominator and solving for $ v $: $$ \frac{1}{v} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} $$ $$ v = 30 \text{ cm} $$
Nature of the image:The image formed by a convex lens, where the object distance is less than the focal length, is virtual, erect, and magnified. The image position is at $ v = 30 \text{ cm} $ to the right of the lens.
Step 2: Placing the concave mirror
The image now acts as a virtual object for the concave mirror. To ensure the concave mirror forms the final image back at the object's original position:
The object (now the image from the lens) should be placed on the concave mirror's center of curvature for it to be imaged back at the same point.
Given:
Radius of curvature of the concave mirror, $ R = 20 \text{ cm }$
The center of curvature is thus $ 20 \text{ cm } $ in front of the mirror.
Conclusion:To make the original position of the object coincide with the final image position using the concave mirror, the mirror should be positioned such that its center of curvature ($ 20 \text{ cm }$ from the mirror) aligns with where the virtual image is formed by the lens, which is $ 30 \text{ cm }$ from the lens position. Adjust the system accordingly to match these distances, factoring in the lens and mirror separation.
What type of reflection of light takes place from: (a) a rough surface? (b) a smooth surface?
(a) Diffuse Reflection: For a rough surface, the reflected light rays scatter in all directions. This occurrence is known as diffuse reflection.
(b) Specular Reflection: With a smooth surface, light reflects at the same angle at which it arrives. Reflected light rays here travel in the same direction. This phenomenon is called specular reflection.
Bouncing back of light from a rough surface is an example of:
A. diffused reflection
B. regular reflection
C. disturbed reflection
D. normal reflection
The correct answer is A. diffused reflection.
When light bounces back from a rough surface, it scatters in multiple directions. This type of scattering is known as diffused reflection, distinguishing it from the smooth, mirror-like regular reflection seen with shiny surfaces.
If a line segment of length $10 \mathrm{~cm}$ is reflected in a line of symmetry (mirror), then the reflection is a line of length $\qquad$
A) $5 \mathrm{~cm}$
B) $10 \mathrm{~cm}$
C) $15 \mathrm{~cm}$
D) $20 \mathrm{~cm}$
The correct answer is B. $10 \mathrm{~cm}$.
When a line segment is reflected across a line of symmetry or a mirror line, the length of the segment remains unchanged. This is because reflections preserve distances between points in the object and its image. Therefore, the reflected segment retains the same length as the original segment.
Consequently, if the original segment is $10 \mathrm{~cm}$ long, the reflected segment will also be $10 \mathrm{~cm}$ long.
A ray 'PQ' of light is incident on the face AB of a glass prism ABC (as shown in the figure) and emerges out of the face AC. Trace the path of the ray. Show that angle i + angle e = angle A + angle δ where δ and e denote the angle of deviation and angle of emergence respectively.
Plot a graph showing the variation of the angle of deviation as a function of the angle of incidence. State the condition under which δ is minimum.
Find out the relation between the refractive index (μ) of the glass prism and angle A for the case when the angle of the prism (A) is equal to the angle of minimum deviation (δm). Hence obtain the value of the refractive index for an angle of prism A = 60°.
Tracing the Path of the Ray:
A ray 'PQ' of light is incident on the face AB of a glass prism ABC. It undergoes refraction at the interface AB and emerges out at the interface AC. Let $e$ be the angle of emergence and $\delta_2$ be the angle of deviation at point Q, as shown in the figure.
Using geometry, at point P, we have: $$ i = \delta_1 + r_1 \quad \text{therefore} \quad \delta_1 = i - r_1 $$
At point Q, we have: $$ e = \delta_2 + r_2 \quad \text{therefore} \quad \delta_2 = e - r_2 $$
The total deviation $\delta$ suffered by the incident ray is equal to $\delta_1 + \delta_2$ $$ \begin{array}{l} \delta = \delta_1 + \delta_2 \ = \left(i - r_1\right) + \left(e - r_2\right) \ = (i + e) - \left(r_1 + r_2\right) \quad \dots (i) \end{array} $$
In quadrilateral POQA, the sum of all four angles is $360^\circ$: $$ P + O + Q + A = 360^\circ $$
Since both $P$ and $Q$ are right angles: $$ \begin{array}{l} \text{P + Q = 180°} \ \text{O + A = 180°} \quad \dots(ii) \end{array} $$
In triangle POQ: $$ O + r_1 + r_2 = 180^\circ \quad \dots(iii) $$
Comparing equations (ii) and (iii), we get: $$ A = r_1 + r_2 $$
Substituting this value in equation (i): $$ \begin{array}{l} \delta = i + e - A \ \delta + A = i + e \quad \dots(iv) \end{array} $$
So, the angle of deviation produced by a prism depends upon the angle of incidence, the refracting angle of the prism, and the material of the prism.
Minimum Deviation Condition:
When the prism is in the position of minimum deviation: $$ i = e \quad \text{and} \quad r_1 = r_2 $$
From equation (iv): $$ \delta_m + A = i + i \ i = \frac{\delta_m + A}{2} $$
The refractive index $\mu$ can be calculated by: $$ \mu = \frac{\sin \left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin \left(\frac{A}{2}\right)} $$
For the case when ( A = \delta_{m} ): $$ \begin{array}{l} \mu = \frac{\sin \left(\frac{A + A}{2}\right)}{\sin \left(\frac{A}{2}\right)} \ = \frac{\sin A}{\sin \left(\frac{A}{2}\right)} \ = \frac{2 \sin \left(\frac{A}{2}\right) \cdot \cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)} \ = 2 \cos \left(\frac{A}{2}\right) \end{array} $$
Given ( A = 60^\circ ): $$ \mu = 2 \cos \left(\frac{60^\circ}{2}\right) = 2 \cos 30^\circ \ = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} $$
Graph:
To plot a graph showing the variation of the angle of deviation $\delta$ as a function of the angle of incidence $i$, we would typically observe that the graph is U-shaped, with the minimum deviation occurring at symmetric incident and emergent angles.
Conclusion:
The angle of deviation ($\delta$) depends on the angles of incidence and the emergent angle ($e$) and the prism’s material. When the ray experiences minimum deviation, the incident and emergent angles are equal. The refractive index ($\mu$) can be determined using the given equations for minimum deviation conditions.
The object through which light passes partially is called an object.
a transparent
a translucent
an opaque
The correct option is B: a translucent.
Light passes partially through a translucent object, which means that objects on the other side are not seen clearly. Some common examples of translucent objects include frosted glass and oil paper.
A ray light is incident at the glass-water interface at an angle $i$, it emerges finally parallel to the surface of the water, then the value of $\mu_{g}$ would be
A $(4/3) \sin i$
B $1/\sin i$
C $4/3$
D 1
The correct answer is B:
$$ \frac{1}{\sin i} $$
Explanation:
When a ray of light is incident at the glass-water interface at an angle $i$ and emerges parallel to the surface of the water, the value of $\mu_{g}$ can be determined as follows:
At the first interface, we have:
$$ \frac{\sin i}{\sin r} = \frac{\mu_{\omega}}{\mu_{\mathrm{g}}} = \frac{4}{3 \mu_{\mathrm{g}}} \tag{1} $$
At the second interface, we observe:
$$ \sin r = \frac{1}{\mu_{\omega}} = \frac{3}{4} \tag{2} $$
By substituting the value from equation (2) into equation (1), we get: $$ \frac{4 \sin i}{3} = \frac{4}{3 \mu_{g}} $$
Solving for $\mu_{g}$, we find: $$ \mu_{g} = \frac{1}{\sin i} $$
Thus, the refractive index $\mu_{g}$ is $\boxed{\frac{1}{\sin i}}$.
A lens behaves as a convergent lens in air and a diverging lens in water. The refractive index of the material of the lens is:
A. Equal to unity
B. Equal to 4/3
C. Between unity and 4/3
D. Greater than 4/3
The correct option is C: Between unity and $\frac{4}{3}$.
To understand why, let's break down the refractive indices:
$$ \mu_{a} < \mu_{1} < \mu_{w}, $$ where:
$\mu_{a}$ represents the refractive index of air, which is 1.
$\mu_{w}$ represents the refractive index of water, which is $\frac{4}{3}$.
For the lens to act as a convergent lens in air and a diverging lens in water, the refractive index of the lens material ($\mu_{1}$) must lie between the refractive indices of air and water, i.e., it must satisfy the condition:
$$ 1 < \mu_{1} < \frac{4}{3}. $$
Thus, the refractive index of the lens material is between unity and $\frac{4}{3}$.
What is/are the condition(s) to achieve total internal reflection?
Light should travel from denser medium to rarer medium
Angle of incidence should be greater than critical angle
Light should travel from rarer medium to denser medium
Both (a) and (b)
The correct answer is D: Both (a) and (b).
Total internal reflection is a phenomenon where light is completely reflected back into the same medium. This occurs under the following conditions:
Light must travel from a denser medium to a rarer medium.
The angle of incidence must be greater than the critical angle.
The critical angle is the angle of incidence at which the angle of refraction becomes $ 90^{\circ} $.
A light ray strikes a mirror normally. What is the angle of reflection?
A. $180^{\circ}$
B. $90^{\circ}$
C. $0^{\circ}$
D. $45^{\circ}$
The correct option is C: $0^\circ$.
The angle of incidence is defined as the angle between the incident ray and the normal to the surface. When a light ray strikes the mirror normally, it means that the incident ray is perpendicular to the mirror surface. Therefore, the angle of incidence is $0^\circ$.
From the first law of reflection:
$$ \text{Angle of incidence} = \text{Angle of reflection} $$
Thus, the angle of reflection is also $0^\circ$.
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