JEE Physics 2025 Question with Solution

Given: A small mirror of mass $m$ is suspended by a massless thread of length $l$. A short laser pulse of energy $E$ falls normally on the mirror.

Find: The small angular deflection $\theta$ of the thread.

For complete reflection, the force due to the laser beam is

$$F = \frac{2P}{c} = \frac{2}{c} \frac{dE}{dt}$$

Hence, the impulse delivered to the mirror is

$$m(V-0) = \int F \, dt = \frac{2}{c} \int dE = \frac{2E}{c}$$

So the speed acquired by the mirror is

$$V = \frac{2E}{mc}$$

At the maximum deflection, the initial kinetic energy converts into gravitational potential energy:

$$\frac{1}{2} mV^2 = mgl(1-\cos\theta)$$

For small angle deflection,

$$1-\cos\theta \approx \frac{\theta^2}{2}$$

Therefore,

$$\frac{1}{2} mV^2 = mgl\left(\frac{\theta^2}{2}\right)$$

Substituting $V = \frac{2E}{mc}$,

$$\frac{1}{2} m \left(\frac{4E^2}{m^2c^2}\right) = mgl\left(\frac{\theta^2}{2}\right)$$ $$\frac{2E^2}{mc^2} = \frac{mgl\theta^2}{2}$$ $$\theta^2 = \frac{4E^2}{m^2c^2gl}$$ $$\theta = \frac{2E}{mc\sqrt{gl}}$$

Therefore, the small angular deflection is $\frac{2E}{mc\sqrt{gl}}$. The correct option is D.

Given: A mirror of mass $m$ hangs as a pendulum of length $l$, and a laser pulse of energy $E$ is reflected normally from it.

Find: The small angular deflection $\theta$.

When light reflects from a mirror, the momentum transferred is

$$\Delta p = \frac{2E}{c}$$

Hence the mirror gets horizontal momentum

$$p = \frac{2E}{c}$$

and corresponding speed

$$v = \frac{p}{m} = \frac{2E}{mc}$$

As the mirror rises through a small angle, all of this kinetic energy changes into potential energy:

$$\frac{1}{2}mv^2 = mgl(1-\cos\theta)$$

Using the small angle approximation,

$$\cos\theta \approx 1-\frac{\theta^2}{2}$$

so

$$1-\cos\theta \approx \frac{\theta^2}{2}$$

Thus,

$$\frac{1}{2}mv^2 = mg l\frac{\theta^2}{2}$$

Now substitute

$$v^2 = \frac{4E^2}{m^2c^2}$$

Then

$$\frac{1}{2}m\left(\frac{4E^2}{m^2c^2}\right) = mg l\frac{\theta^2}{2}$$ $$\frac{2E^2}{mc^2} = mg l\frac{\theta^2}{2}$$ $$\theta^2 = \frac{4E^2}{m^2glc^2}$$

Taking the positive root,

$$\theta = \frac{2E}{mc\sqrt{gl}}$$

Therefore, the correct option is D.