Electromagnetic Waves - Class 12 Physics - Chapter 9 - Notes, NCERT Solutions & Extra Questions
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Notes - Electromagnetic Waves | Class 12 NCERT | Physics
Comprehensive Electromagnetic Waves Class 12 Notes: Key Concepts and Summary
Introduction to Electromagnetic Waves
What are Electromagnetic Waves?
Electromagnetic waves are waves composed of oscillating electric and magnetic fields that travel through space. These waves are capable of propagating without the need for a medium, making them unique compared to other types of waves.
Discovery and Historical Context
James Clerk Maxwell, a prominent physicist, formulated a set of equations that predicted the existence of electromagnetic waves. Heinrich Hertz later confirmed these predictions through experiments, laying the groundwork for modern electromagnetic wave theory.
Maxwell's Equations and Electromagnetic Waves
The Significance of Maxwell's Equations
Maxwell's Equations are fundamental in understanding electromagnetism. They encapsulate how electric and magnetic fields interact and propagate. These equations are:
Gauss's Law for Electricity
Gauss's Law for Magnetism
Faraday's Law of Induction
Ampere-Maxwell Law
Role of Displacement Current
Maxwell introduced the concept of displacement current to resolve inconsistencies in Ampere's Law. This additional term accounts for the changing electric field, ensuring the magnetic field is consistent regardless of the surface used in calculations.
Production and Propagation of Electromagnetic Waves
How Electromagnetic Waves are Produced
Electromagnetic waves are generated by accelerated charges, such as electrons. An oscillating electric charge produces oscillating electric and magnetic fields, which propagate as electromagnetic waves.
Propagation in Space
The speed of electromagnetic waves in a vacuum is denoted by ( c ), approximately ( 3 \times 10^8 ) metres per second. In materials, the speed is influenced by the medium's permeability and permittivity:
[ v = \frac{1}{\sqrt{\mu \varepsilon}} ]
Electromagnetic Spectrum
Electromagnetic waves are classified based on their wavelengths and frequencies, forming the electromagnetic spectrum.
Radio Waves
Radio waves have the longest wavelengths and are used for communication, such as radio and TV broadcasting.
Microwaves
Microwaves are used in radar systems, communication, and microwaving food. They possess shorter wavelengths than radio waves.
Infrared Waves
Infrared waves are produced by hot objects and molecules. They find applications in thermal imaging, remote controls, and even cooking.
Visible Light
Visible light is detected by the human eye. It ranges from 400 nm to 700 nm in wavelength and is essential for human sight.
Ultraviolet Rays
Ultraviolet rays have shorter wavelengths than visible light. They are used in sterilisation and have various industrial applications.
X-rays
X-rays are used extensively in medical imaging. They possess high energy and can penetrate many materials.
Gamma Rays
Gamma rays have the shortest wavelengths and highest frequencies. They are produced by nuclear reactions and are used in cancer treatment.
Applications and Technological Importance
Communication Systems
Electromagnetic waves play a crucial role in communication technologies, from radio and TV broadcasting to mobile phones and Wi-Fi.
Medical Uses
X-rays and gamma rays are integral in medical diagnostics and treatments, such as cancer radiotherapy. Ultraviolet rays are used for sterilising equipment and water.
Everyday Applications
Microwaves are used in ovens to heat food, and infrared sensors are utilised in various remote control devices.
Mathematical Representation
Equations Governing Electromagnetic Waves
Electromagnetic waves can be described mathematically with sinusoidal functions:
[ \begin{aligned} E_x & = E_0 \sin(kz - \omega t) \ B_y & = B_0 \sin(kz - \omega t) \end{aligned} ]
Where ( k ) is the wave number and ( \omega ) is the angular frequency.
Example Calculations
Finding Magnetic Field from Electric Field:
[ B = \frac{E}{c} ]
Calculating Wavelength and Frequency:
[ \lambda = \frac{2\pi}{k} ]
[ v = f \lambda = c ]
Conclusion
Summary of Key Points
Electromagnetic waves encompass a wide range of frequencies, each with unique properties and applications. Understanding these waves and their behaviours is crucial for advancements in technology, communication, and medical fields.
Electromagnetic Wave Structure
graph LR
EField["Electric Field (E)"] --> Propagation[Propagation Direction]
MField["Magnetic Field (B)"] --> Propagation
EField --> Perpendicular
MField --> Perpendicular
Perpendicular -- "90 degrees" --> Propagation
Note
Understanding electromagnetic waves is foundational for students in Class 12 as these concepts bridge theoretical physics and practical applications, fostering innovation and comprehension of the natural world.
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Notes - Flashcards - Electromagnetic Waves | Class 12 NCERT | Physics
NCERT Solutions - Electromagnetic Waves | NCERT | Physics | Class 12
Figure 8.5 shows a capacitor made of two circular plates each of radius $12 \mathrm{~cm}$, and separated by $5.0 \mathrm{~cm}$. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.
(a) Calculate the capacitance and the rate of change of potential difference between the plates.
(b) Obtain the displacement current across the plates.
(c) Is Kirchhoffs first rule (junction rule) valid at each plate of the capacitor? Explain.
Problem Breakdown
Radius $r$ of the capacitor plates: $12 \text{ cm} = 0.12 \text{ m}$
Separation $d$ between the plates: $5.0 \text{ cm} = 0.05 \text{ m}$
Charging current $I = 0.15 \text{ A}$
Part (a)
Calculate the Capacitance ( C ):
The capacitance ( C ) of a parallel plate capacitor is given by: $$ C = \frac{\epsilon_0 A}{d} $$ where:
( \epsilon_0 ) (permittivity of free space) ( \approx 8.85 \times 10^{-12} \text{ F/m} )
( A ) (area of the plates) ( = \pi r^2 )
( d ) (separation between the plates)
Given area of the plates, (A = \pi (0.12)^2 \text{ m}^2)
Given separation between the plates, ( d = 0.05 \text{ m} )
Calculate the Rate of Change of Potential Difference ((dV/dt)):
We have the relation between the current ( I ), capacitance ( C ), and the rate of change of potential difference: $$ I = C \frac{dV}{dt} $$ Rearranging, we get: $$ \frac{dV}{dt} = \frac{I}{C} $$
Part (b)
Displacement Current ( I_d ) across the plates:
The displacement current ( I_d ) is given by: $$ I_d = C \frac{dV}{dt} $$
Given the current is ( 0.15 \text{ A} ), the displacement current must be equal to the conduction current.
Part (c)
Validity of Kirchhoff's First Rule:
Kirchhoff's First Rule (junction rule) states that the sum of currents entering a junction must equal the sum leaving. In the case of the capacitor, the conduction current entering the positive plate should equal the displacement current leaving it.
Now, calculate the required values:
Part (a) - Capacitance Calculation
$$ C = \frac{\epsilon_0 \cdot \pi (0.12)^2 }{0.05} $$
Part (a) - Rate of Change of Potential Difference
$$ \frac{dV}{dt} = \frac{I}{C} $$
Part (b) - Displacement Current
Given, ( I = 0.15 \text{ A} ), we know that ( I_d = 0.15 \text{ A} )
Following these formulae sequentially:
Capacitance ( C ) Calculation: Let's calculate the capacitance: $$ C = \frac{8.854 \times 10^{-12} \cdot 3.14 \cdot (0.12)^2 }{0.05} $$
Rate of change of potential difference:$$ \frac{dV}{dt} = \frac{0.15}{C} $$
Displacement Current:Displacement current will be ( 0.15 \text{ A} ).
Let's calculate these step by step:
Part (a)
Capacitance:
[ A = \pi r^2 = \pi (0.12)^2 \ A = 0.0452 , m^2 \ C = \frac{\epsilon_0 \cdot A}{d} = \frac{8.85 \times 10^{-12} \times 0.0452}{0.05} ]
Rate of Change of Potential Difference:
[ \frac{dV}{dt} = \frac{0.15}{C} ]
Part (b)
Displacement Current:
[ I_d = 0.15 , \text{A} ]
Computing the values using a precise tool to ensure correctness:
1. Capacitance \(C\):
\[ C = 8.002 \times 10^{-12} \, F \]
2. Rate of Change of Potential Difference:
\[ \frac{dV}{dt} = 1.875 \, \text{V/s} \]
3. Displacement Current:
$$ \(I_d = 0.15 \, \text{A}\) $$
Summary:
a) The capacitance of the capacitor is \( 8.002 \times 10^{-12} \, F \) and the rate of change of potential difference is \( 1.875 \, \text{V/s} \).
b) The displacement current across the plates is \( 0.15 \, \text{A} \).
c) Validity of Kirchhoff's First Rule:
- Kirchhoff's first rule is valid because the displacement current \( I_d = 0.15 \, \text{A} \) ensures that the current entering one plate equals the current leaving the other, maintaining current continuity.
A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius $R=6.0 \mathrm{~cm}$ has a capacitance $C=100 \mathrm{pF}$. The capacitor is connected to a $230 \mathrm{~V}$ ac supply with a (angular) frequency of $300 \mathrm{rad} \mathrm{s}^{-1}$.
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of $\mathbf{B}$ at a point $3.0 \mathrm{~cm}$ from the axis between the plates.
The amplitude of the magnetic field ( B ) at a point 3.0 cm from the axis between the plates is:
[ B = \frac{4 \pi \times 10^{-7} \text{H/m} \times 4.88 \times 10^{-6} \text{A} \times 0.03 \text{m}}{2 \pi \times (0.06 \text{m})^2} ]
This simplifies to:
[ B = \frac{8.13333 \times 10^{-12} \text{A} \cdot \text{H}}{\text{m}^2} ]
To put it in more conventional units:
[ B \approx 8.1 \times 10^{-12} \ \text{T} \ ( \text{Tesla} ) ]
Hence, the amplitude of the magnetic field at a point 3.0 cm from the axis between the capacitor plates is ( 8.1 \times 10^{-12} \ \text{T} ).
What physical quantity is the same for X-rays of wavelength $10^{-10} \mathrm{~m}$, red light of wavelength $6800 \AA$ and radiowaves of wavelength $500 \mathrm{~m}$ ?
The physical quantity that is the same for X-rays, red light, and radiowaves is the speed of propagation in a vacuum. This speed is the speed of light, which is $3 \times 10^8$ m/s.
This is a fundamental property of all electromagnetic waves, regardless of their wavelength or frequency.
A plane electromagnetic wave travels in vacuum along $z$-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is $30 \mathrm{MHz}$, what is its wavelength?
The wavelength of an electromagnetic wave with a frequency of $30 \text{ MHz}$ is approximately 9.99 meters.
Summary
Propagation Direction: $z$-direction
Electric Field Direction: $x$-direction
Magnetic Field Direction: $y$-direction
Frequency: $30 \text{ MHz}$
Wavelength: $9.99 \text{ meters}$
Hence, $\lambda \approx 9.99 \text{ meters}$.
A radio can tune in to any station in the $7.5 \mathrm{MHz}$ to $12 \mathrm{MHz}$ band. What is the corresponding wavelength band?
The radio can tune into wavelengths corresponding to frequencies in the $7.5 \mathrm{MHz}$ to $12 \mathrm{MHz}$ band.
At $7.5 \mathrm{MHz}$, the corresponding wavelength is 39.97 meters.
At $12 \mathrm{MHz}$, the corresponding wavelength is 24.98 meters.
Thus, the wavelength band for the radio is from 24.98 meters to 39.97 meters.
A charged particle oscillates about its mean equilibrium position with a frequency of $10^{9} \mathrm{~Hz}$. What is the frequency of the electromagnetic waves produced by the oscillator?
The frequency of the electromagnetic waves produced by an oscillator is the same as the frequency of the oscillating charged particle. Hence, if a charged particle oscillates with a frequency of $10^9 \text{ Hz}$, the frequency of the electromagnetic waves produced will also be $10^9 \text{ Hz}$.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_{0}=510 \mathrm{nT}$. What is the amplitude of the electric field part of the wave?
The amplitude of the electric field part of the wave ( E_0 ) is:
$$ E_0 = 153 , \text{V/m} $$
So, the amplitude of the electric field part of the wave is 153 V/m.
Suppose that the electric field amplitude of an electromagnetic wave is $E_{0}=120 \mathrm{~N} / \mathrm{C}$ and that its frequency is $v=50.0 \mathrm{MHz}$. (a) Determine, $B_{0}, \omega, k$, and $\lambda$. (b) Find expressions for $\mathbf{E}$ and $\mathbf{B}$.
Part (a): Calculated Values
Magnetic Field Amplitude ( B_0 )
[ B_0 = \frac{E_0}{c} = \frac{120 , \mathrm{N/C}}{3 \times 10^8 , \mathrm{m/s}} = 4 \times 10^{-7} , \mathrm{T} ]
Angular Frequency ( \omega )
[ \omega = 2 \pi v = 2 \pi \times 50 \times 10^6 , \mathrm{Hz} \approx 3.14 \times 10^8 , \mathrm{rad/s} ]
Wavelength ( \lambda )
[ \lambda = \frac{c}{v} = \frac{3 \times 10^8 , \mathrm{m/s}}{50 \times 10^6 , \mathrm{Hz}} = 6 , \mathrm{m} ]
Wave Number ( k )
[ k = \frac{2 \pi}{\lambda} = \frac{2 \pi}{6 , \mathrm{m}} \approx 1.047 , \mathrm{rad/m} ]
Part (b): Expressions for ( \mathbf{E} ) and ( \mathbf{B} )
Considering the wave is traveling along the ( z )-axis, the electric and magnetic fields can be expressed as:
Electric Field ( \mathbf{E} )
[ \mathbf{E} = E_0 \sin(kz - \omega t) \hat{x} = 120 \sin(1.047z - 3.14 \times 10^8 t) , \hat{x} , \mathrm{N/C} ]
Magnetic Field ( \mathbf{B} )
[ \mathbf{B} = B_0 \sin(kz - \omega t) \hat{y} = 4 \times 10^{-7} \sin(1.047z - 3.14 \times 10^8 t) , \hat{y} , \mathrm{T} ]
Summary of Results:
Magnetic Field Amplitude: ( B_0 = 4 \times 10^{-7} , \mathrm{T} )
Angular Frequency: ( \omega \approx 3.14 \times 10^8 , \mathrm{rad/s} )
Wave Number: ( k \approx 1.047 , \mathrm{rad/m} )
Wavelength: ( \lambda = 6 , \mathrm{m} )
Electric Field Expression: [ \mathbf{E} = 120 \sin(1.047z - 3.14 \times 10^8 t) , \hat{x} , \mathrm{N/C} ]
Magnetic Field Expression: [ \mathbf{B} = 4 \times 10^{-7} \sin(1.047z - 3.14 \times 10^8 t) , \hat{y} , \mathrm{T} ]
The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula $E=h \nu$ (for energy of a quantum of radiation: photon) and obtain the photon energy in units of $\mathrm{eV}$ for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?
Here are the calculated photon energies for different parts of the electromagnetic spectrum, expressed in electronvolts (eV):
Radio Waves
Frequency: ( 10^8 , \text{Hz} )
Photon Energy: $4.136 \times 10^{-7} , \text{eV}$
Microwaves
Frequency: ( 10^{11} , \text{Hz} )
Photon Energy: $4.136 \times 10^{-4} , \text{eV}$
Infrared
Frequency: ( 10^{13} , \text{Hz} )
Photon Energy: $0.04136 , \text{eV}$
Visible Light
Frequency: ( 5.5 \times 10^{14} , \text{Hz} )
Photon Energy: $2.275 , \text{eV}$
Ultraviolet
Frequency: ( 10^{16} , \text{Hz} )
Photon Energy: $41.36 , \text{eV}$
X-rays
Frequency: ( 10^{19} , \text{Hz} )
Photon Energy: $41.36 , \text{k}\text{eV}$
Gamma Rays
Frequency: ( 10^{22} , \text{Hz} )
Photon Energy: $41.36 , \text{MeV}$
Relationship Between Photon Energies and Sources
Radio Waves: These have very low photon energies and are generated by the acceleration of electrons in antennas (e.g., broadcasting antennas for radio and TV).
Microwaves: With higher photon energies than radio waves, they are produced by special vacuum tubes like klystrons and magnetrons, useful in radar systems and microwave ovens.
Infrared: These photons are generated by hot objects and molecules and have energies high enough to cause molecular vibrations, hence their association with heat.
Visible Light: Produced by electronic transitions in atoms and molecules, visible light photons have moderate energy levels that correspond to chemical bond energies, making them detectable by the human eye.
Ultraviolet: These photons have enough energy to cause excitation and ionization of atoms. UV light sources include the Sun and special UV lamps.
X-rays: High-energy photons produced typically by electron acceleration and deceleration processes in X-ray tubes. They are used for medical imaging and materials inspection.
Gamma Rays: With the highest photon energies, gamma rays are generated by nuclear reactions and radioactive decay. They are used in medical treatments (e.g., cancer therapy) and by astrophysical phenomena.
The increasing photon energy from radio waves to gamma rays correlates with increasingly energetic processes needed to generate these electromagnetic waves.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10} \mathrm{~Hz}$ and amplitude $48 \mathrm{~V} \mathrm{~m}^{-1}$.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the $\mathbf{E}$ field equals the average energy density of the $\mathbf{B}$ field. $\left[c=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\right.$.]
Part (a): Wavelength of the Wave
The wavelength, $\lambda$, is given by:
$$ \lambda = \frac{c}{f} $$
Plugging in the given values:
$$ \lambda = \frac{3 \times 10^8 , \text{m/s}}{2 \times 10^{10} , \text{Hz}} = 0.015 , \text{m} $$
So, the wavelength of the wave is 0.015 meters.
Part (b): Amplitude of the Oscillating Magnetic Field
The amplitude of the oscillating magnetic field, $B_0$, is given by:
$$ B_0 = \frac{E_0}{c} $$
Given:
$E_0 = 48 , \text{V/m}$
$c = 3 \times 10^8 , \text{m/s}$
Let's calculate: amplitude of magnetic field = 48 / (3 * 10^8)
$$ B_0 = \frac{48}{3 \times 10^8} = 1.6 \times 10^{-7} , \text{T} $$
So, the amplitude of the oscillating magnetic field is 1.6 × 10^{-7} T.
Part (c): Average Energy Density Equality
The average energy densities of the electric field and magnetic field are given by:
$$ u_E = \frac{1}{2} \varepsilon_0 E_0^2 $$
$$ u_B = \frac{1}{2} \frac{B_0^2}{\mu_0} $$
We need to show: $$ u_E = u_B $$
Using the relationship $B_0 = \frac{E_0}{c}$:
$$ u_E = \frac{1}{2} \varepsilon_0 E_0^2 $$
$$ u_B = \frac{1}{2} \frac{\left(\frac{E_0}{c}\right)^2}{\mu_0} = \frac{1}{2} \frac{E_0^2}{c^2 \mu_0} $$
Since $c^2 = \frac{1}{\mu_0 \varepsilon_0}$, we have:
$$ c^2 \mu_0 = \frac{1}{\varepsilon_0} \ \varepsilon_0 = \frac{1}{c^2 \mu_0} $$
Hence,
$$ u_B = \frac{1}{2} \varepsilon_0 E_0^2 $$
Thus, we have shown that:
$$ u_E = u_B $$
So, the average energy density of the electric field equals the average energy density of the magnetic field.
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A radar has a power of $1 \mathrm{KW}$ and is operating at a frequency of $10 \mathrm{GHz}$. It is located on a mountain top of height $500 \mathrm{~m}$. The maximum distance up to which it can detect object located on the surface of the earth is
[Radius of earth = 6.4 \times 10^{6} \mathrm{~m}].
A) $80 \mathrm{~km}$
B) $16 \mathrm{~km}$
C) $40 \mathrm{~km}$
D) $64 \mathrm{~km}$
To calculate the maximum distance at which a radar located on a mountain can detect an object on the Earth's surface, we utilize the formula:
$$ \text{Maximum distance} = \sqrt{(h+R)^2 - R^2} $$
Where:
$ h = 500 , \text{m} $ (height of the mountain)
$ R = 6.4 \times 10^6 , \text{m} $ (radius of the Earth)
Simplifying the above expression: $$ \sqrt{(h+R)^2 - R^2} = \sqrt{R^2 + 2Rh + h^2 - R^2} = \sqrt{2Rh + h^2} $$
Approximating further under the assumption that $ h $ is much smaller than $ R $, the term $ h^2 $ is negligible compared to $ 2Rh $: $$ \text{Maximum distance} \approx \sqrt{2Rh} $$
Substituting the values for $ h $ and $ R $: $$ = \sqrt{2 \times 6.4 \times 10^6 \times 500} $$
To simplify calculation, note that: $$ 2 \times 6.4 \times 10^6 \times 500 = 6.4 \times 10^6 \times 1000 = 6.4 \times 10^9 $$
Which means: $$ \sqrt{6.4 \times 10^9} $$
Can be approximated as: $$ = 80 \times 10^3 , \text{m} = 80 , \text{km} $$
Thus, the correct option is A) $80 , \text{km}$.
Interference of light provides evidence for the fact that
A. light travels with a high speed
B. light is electromagnetic in nature
C. light is a wave phenomenon
D. light is a transverse wave
The correct answer is C. Light is a wave phenomenon.
Interference of light occurs when two or more waves superimpose to form a resultant wave of greater or lesser amplitude. This phenomenon provides direct evidence that light behaves as a wave. Such behavior is characteristic of waves, confirming that light indeed exhibits wave properties. Hence, option C is the correct choice.
The correct sequence of frequency for electromagnetic radiations is:
A) Radio waves > Microwaves > UV rays > X-rays B) X-rays > UV rays > Microwaves > Radio waves C) Radio waves > Microwaves > X-rays > UV rays D) UV rays > X-rays > Radio waves > Microwaves
The correct answer is B) X-rays > UV rays > Microwaves > Radio waves.
The sequence is based on the frequency of electromagnetic radiations, which ranges as follows:
- X-rays: from $10^{16}$ to $10^{20}$ Hz
- UV rays: from $10^{14}$ to $10^{16}$ Hz
- Microwaves: from $10^{8}$ to $10^{10}$ Hz
- Radio waves: from $10^{4}$ to $10^{8}$ Hz
This order clearly shows that X-rays have the highest frequency, followed by UV rays, Microwaves, and lastly Radio waves, confirming the sequence in option B.
Which of the following radiations has the least wavelength?
A) $\gamma$-rays
B) $\beta$-rays
C) $\alpha$-rays
D) $X$-rays
Solution
The correct option is A - $\gamma$-rays.
$\gamma$-rays have the least wavelength among the given options.
An electromagnetic wave going through vacuum is described by $\mathrm{E}=E_{o} \sin (k x-\omega t) ; B=B_{0} \sin (k x-\omega t)$. Which of the following equations is true?
(A) $E_{0} k=B_{0} \omega$
(B) $E_{o} \omega=B_{0} k$
(C) $\mathrm{E}_{0} \mathrm{~B}_{\mathrm{o}}=\omega \mathrm{k}$
(D) None of these
The correct answer is (A) $E_{0} k=B_{0} \omega$
To understand why this is the correct option, we can recall that in a vacuum, the ratio of the magnitudes of the electric field $E_0$ and the magnetic field $B_0$ of an electromagnetic wave is equal to the speed of light $c$. Hence, we have the relation: $$ \frac{E_0}{B_0} = c $$ Additionally, the wave number $k$ and the angular frequency $\omega$ are related to the properties of the wave: $$ k = \frac{2\pi}{\lambda} \quad \text{and} \quad \omega = 2\pi f $$ Substituting the relation between $E_0$, $B_0$, and $c$ into the wave equations, and knowing that $c = \frac{\omega}{k}$, we can derive the relationship: $$ E_{0} k = B_{0} \omega $$ Thus, the correct option is (A).
In Australia, people are suffering from skin cancer due to __________ from the sun.
A) gamma rays
B) ultraviolet rays
C) infrared rays
D) visible rays
The correct answer is B) ultraviolet rays.
Australia experiences one of the highest rates of skin cancer globally. This alarming prevalence is predominantly due to exposure to UV radiation found in sunlight. The ozone layer plays a crucial role in blocking UV radiation, but factors like air pollution have led to its depletion. This reduction in ozone allows more UV rays to penetrate the Earth's atmosphere, significantly elevating the risk of skin-related conditions, including skin cancer.
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