# Wave Optics - Class 12 - Physics

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## Extra Questions - Wave Optics | NCERT | Physics | Class 12

The number of photons emitted by a $5 \mathrm{~mW}$ laser source operating at $632.8 \mathrm{~nm}$ is $\qquad$ $\times 10^{16}$.

To calculate the number of photons emitted by the laser, first, determine the **energy of each photon** using the formula:
$$
E = \frac{hc}{\lambda}
$$
where

$ h = 6.63 \times 10^{-34} \, \text{Js} $ is the

**Planck constant**,$ c = 3 \times 10^8 \, \text{m/s} $ is the

**speed of light**, and$ \lambda = 632.8 \times 10^{-9} \, \text{m} $ is the

**wavelength**of the laser light.

Plugging in the values: $$ E = \frac{(6.63 \times 10^{-34} \text{ Js}) \times (3 \times 10^8 \text{ m/s})}{632.8 \times 10^{-9} \text{ m}} = 3.14 \times 10^{-19} \text{ J} $$

The **total energy emitted per second** by the laser is $ 5 \text{ mW} = 5 \times 10^{-3} \text{ J/s} $.

Now, calculate the **number of photons** emitted per second by dividing the total energy per second by the energy per photon:
$$
\text{Number of photons per second} = \frac{5 \times 10^{-3} \text{ J}}{3.14 \times 10^{-19} \text{ J}} = 1.6 \times 10^{16}
$$

Therefore, the number of photons emitted by the laser source per second is approximately $1.6 \times 10^{16}$.

A simple harmonic wave is represented by the relation $y(x, t) = a_{0} \sin 2\pi \left(vt - \frac{x}{\lambda})$. If the maximum particle velocity is three times the wave velocity, the wavelength $\lambda$ of the wave is:

(A) $\frac{\pi a_{0}}{3}$

(B) $\frac{2\pi a_{0}}{3}$

(C) $\pi a_{0}$

(D) $\frac{\pi a_{0}}{2}$

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A standing wave is represented by $y = a \sin(0.05x) \cos(100t)$, where $t$ is in $s$ and $x$ is in $\mathrm{m}$. Then the velocity of the constituent wave is

(A) $10^{2} \mathrm{~m/s}$

(B) $10^{5} \mathrm{~m/s}$

(C) $10^{3} \mathrm{~m/s}$

(D) $2 \times 10^{3} \mathrm{~m/s}$