1. Momentum carried by the laser pulse

A pulse of electromagnetic radiation of energy $E$ carries momentum

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

Since the pulse falls normally on the mirror and gets reflected, the momentum change of the light is reversed.

So, momentum transferred to the mirror is

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

Hence the mirror gets an instantaneous speed $v$ such that

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

or

$v = \frac{2E}{mc}.$

2. Conversion of kinetic energy into gravitational potential energy

After receiving the impulse, the mirror starts moving and rises to a maximum angular deflection $\theta$.

At the highest point, all initial kinetic energy converts into gravitational potential energy:

$\frac{1}{2}mv^2 = mg h,$

where for small angular displacement,

$h = l(1-\cos\theta).$

For small $\theta$,

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

so

$h \approx \frac{l\theta^2}{2}.$

Thus,

$\frac{1}{2}mv^2 = mg\left(\frac{l\theta^2}{2}\right).$

Canceling $\frac{1}{2}m$ from both sides,

$v^2 = gl\theta^2.$

Therefore,

$\theta = \frac{v}{\sqrt{gl}}.$

Substitute $v = \frac{2E}{mc}$:

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

3. Matching with options

$\boxed{\theta = \frac{2E}{m c \sqrt{g l}}}$

This matches Option D.

4. Verification with stored answer

Stored correct answer: D
Derived answer: D

So they agree.