According to the question
$$L =$$  Light source
When the light beams incident normally on the plane mirror, it is reflected back to the point from which it was coming. When the plane mirror is rotated by an angle $$\theta $$, the reflected ray or beam of light must rotate by angle $$2\theta ,$$ from refraction at plane surface theory.

From the figure $$\tan 2\theta = \frac{{BS}}{{SO}} = \frac{y}{x}$$
If the angle is small $$\tan 2\theta \approx 2\theta $$
So, $$2\theta = \frac{y}{x} \Rightarrow \theta = \frac{y}{{2x}}$$

Let refractive index of glass be $$\mu .$$
Let after first refraction, image distance be $$v$$ then
$$\eqalign{ & \frac{\mu }{v} - \frac{1}{\infty } = \frac{{\mu - 1}}{R} \cr & \Rightarrow v = \frac{{\mu R}}{{\mu - 1}} \cr} $$
Now second refraction will take place.
So distance of first image from $$O$$ is
$$\eqalign{ & {u_1} = \frac{{\mu R}}{{\mu - 1}} - R = \frac{R}{{\mu - 1}}\,\,{\text{and image is formed at }}R \cr & \therefore \frac{1}{R} - \frac{{\mu \left( {\mu - 1} \right)}}{R} = \frac{{2\left( {1 - \mu } \right)}}{R} \cr & \Rightarrow {\mu ^2} - 3\mu + 1 = 0 \cr & {\text{So,}}\,\,\mu = \frac{{3 + \sqrt 5 }}{2} \cr} $$

In a prism: $$r + r' = A$$
$$\eqalign{ & \Rightarrow r = A - r' \cr & \therefore r = {60^ \circ } - \left( {10 + {t^2}} \right) = 50 - {t^2} \cr} $$

According to the question

From the above diagram, intensity transmitted through $${P_3}$$
$$\eqalign{ & {I_2} = \frac{{{I_0}}}{2}{\cos ^2}{45^ \circ } \Rightarrow {I_2} = \frac{{{I_0}}}{2} \times {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \cr & \Rightarrow {I_2} = \frac{{{I_0}}}{4} \cr} $$
Similarly, intensity transmitted through $${P_2},$$
$$\eqalign{ & {I_3} = \frac{{{I_0}}}{4}{\cos ^2}{45^ \circ } \Rightarrow {I_3} = \frac{{{I_0}}}{4} \times {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \cr & \Rightarrow {I_3} = \frac{{{I_0}}}{4} \times \frac{1}{2} \Rightarrow {I_3} = \frac{{{I_0}}}{8} \cr} $$

As we know that
Intensity, $$I \propto A$$   (Area exposed)
$$\eqalign{ & \Rightarrow \frac{{{I_2}}}{{{I_1}}} = \left[ {\frac{{{A_2}}}{{{A_1}}}} \right] = \frac{{\frac{{\pi {d^2}}}{4} - \frac{{\frac{{\pi {d^2}}}{4}}}{4}}}{{\frac{{\pi {d^2}}}{4}}} = \frac{3}{4} \cr & \Rightarrow {I_2} = \frac{3}{4}{I_1} \cr} $$
and focal length remains unchanged.

For concave mirror
$$\eqalign{ & \frac{2}{R} = \frac{1}{v} + \frac{1}{u}\,\,{\text{or}}\,\,\frac{2}{{ - R}} = \frac{1}{v} + \frac{1}{{ - u}} \cr & \therefore \frac{1}{v} = \frac{1}{U} - \frac{2}{R} = \frac{{R - 2U}}{{UR}} \cr & {\text{or}}\,\,v = \left[ {\frac{{RU}}{{R - 2U}}} \right] \cr} $$
In spherical mirror, image velocity
$${v_i} = - \left[ {\frac{{{v^2}}}{{{u^2}}}} \right]{v_0} = - {\left[ {\frac{{RU}}{{R - 2U}}} \right]^2}\frac{{{v_0}}}{{{U^2}}} = - {\left[ {\frac{R}{{R - 2U}}} \right]^2}{v_0}$$

$$\eqalign{ & \sin {i_c} = \frac{1}{\mu } = \frac{r}{{\sqrt {{r^2} + {h^2}} }} \cr & {\text{Using}}\,\,h = 12\,cm,\mu = \frac{4}{3} \cr & {\text{We}}\,{\text{get}}\,\,r = \frac{{36}}{{\sqrt 7 }}cm. \cr} $$

A concave mirror forms real and virtual images, whose magnification can be negative or positive depending upon the position of the object. If object is placed between focus and pole the image obtained will be virtual and its magnification will be positive. In all other cases concave mirror forms real images whose magnification will be negative. A convex mirror always forms a virtual image whose magnification will always be positive.

This graph obeys the lens equation
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
where $$f$$ is a positive constant for a given convex lens

As and we know that $${P_{{\text{eq}}}} = {P_1} + {P_2}$$
$$\eqalign{ & \frac{1}{{{F_{{\text{eq}}}}}} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} \cr & \therefore {F_{{\text{eq}}}} = \frac{{{f_1}{f_2}}}{{{f_1} + {f_2}}} \cr & {\text{As}}\,\,P = \frac{1}{F} \cr & \therefore {P_{{\text{eq}}}} = \frac{{{f_1} + {f_2}}}{{{f_1}{f_2}}} \cr} $$