A small mirror of mass $m$ is suspended by a massless thread of length $l$. Then the small angle through which the thread will be deflected when a short pulse of laser of energy $E$ falls normal on the mirror ( $c=$ speed of light in vacuum and $g=$ acceleration due to gravity).
Show Hint
- $\theta=\frac{3 \mathrm{E}}{4 \mathrm{mc} \sqrt{g l}}$
- $\theta=\frac{\mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} l}}$
- $\theta=\frac{\mathrm{E}}{2 \mathrm{mc} \sqrt{\mathrm{gl}}}$
- $\theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{gl}}}$
The Correct Option is D
Approach Solution - 1
2. Change in momentum of the mirror: \[ m(V - 0) = \int F dt = \frac{2}{c} \int dE = \frac{2E}{c} \]
3. Using work-energy theorem: \[ W_g = \Delta K \] \[ -\frac{mg l (1 - \cos \theta)}{l} = \frac{1}{2} m V^2 \] \[ \frac{g l (1 - \cos \theta)}{l} = \frac{1}{2} \left( \frac{4E^2}{m^2 c^2} \right) \] \[ \frac{g l \theta^2}{2} = \frac{2E^2}{m^2 c^2} \] \[ \theta = \frac{2E}{mc \sqrt{gl}} \] Therefore, the correct answer is (4) $\theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{gl}}}$.
Approach Solution -2
A small mirror of mass \( m \) is suspended by a thread of length \( l \). A short laser pulse of energy \( E \) falls normally on the mirror. We are to find the small angular deflection \( \theta \) of the thread due to the momentum imparted by the reflected light.
Concept Used:
When light of energy \( E \) is reflected from a mirror, it transfers momentum to the mirror due to radiation pressure. The momentum imparted to the mirror (for complete reflection) is given by:
\[ \Delta p = \frac{2E}{c} \]Because light reverses direction upon reflection, the momentum change is twice the incident momentum. The mirror, in turn, acquires this momentum. This horizontal momentum produces a deflection in the pendulum, where tension and gravity balance the forces at the maximum deflection.
Step-by-Step Solution:
Step 1: The impulse on the mirror due to reflection of the light pulse is:
\[ p = \frac{2E}{c} \]Step 2: The mirror (mass \( m \)) acquires a horizontal velocity \( v \) due to this impulse:
\[ v = \frac{p}{m} = \frac{2E}{mc} \]Step 3: As the mirror moves, it rises along an arc until all the kinetic energy is converted into potential energy. Applying conservation of energy:
\[ \frac{1}{2} m v^2 = m g l (1 - \cos\theta) \]Step 4: For small deflection, \( \cos\theta \approx 1 - \frac{\theta^2}{2} \). Substituting this approximation:
\[ \frac{1}{2} m v^2 = m g l \left( \frac{\theta^2}{2} \right) \]Step 5: Simplify and substitute \( v = \frac{2E}{mc} \):
\[ v^2 = \frac{4E^2}{m^2 c^2} \] \[ \frac{1}{2} m \left( \frac{4E^2}{m^2 c^2} \right) = m g l \left( \frac{\theta^2}{2} \right) \]Step 6: Simplify the equation to solve for \( \theta \):
\[ \frac{2E^2}{m c^2} = m g l \frac{\theta^2}{2} \] \[ \theta^2 = \frac{4E^2}{m^2 g l c^2} \]Step 7: Taking the square root:
\[ \theta = \frac{2E}{m c \sqrt{g l}} \]Final Computation & Result:
Hence, the small angular deflection of the thread is given by:
\[ \boxed{\theta = \frac{2E}{m c \sqrt{g l}}} \]Final Answer: \( \theta = \dfrac{2E}{m c \sqrt{g l}} \)
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