Consider a ray of light $$PQ$$  incident an the surface $$AB$$  and moves along $$RS,$$  after passing through the prism $$ABC.$$
It is given that the incident ray suffers minimum deviation. Therefore, the ray inside the prism must be parallel to the base $$BC$$ of the prism. From the geometry of the prism and the ray diagram, it is clear that
angle of incidence, $$i = {45^ \circ }$$
angle of refraction $$r = r' = {30^ \circ }$$
angle of emergence, $$e = {45^ \circ }$$
Therefore, minimum deviation suffered by the ray is
$${\delta _{\min }} = i + e - \left( {r + r'} \right) = {90^ \circ } - {60^ \circ } = {30^ \circ }$$
Also we know that
$$\mu = \frac{{\sin \left( {\frac{{A + {\delta _m}}}{2}} \right)}}{{\sin \frac{A}{2}}}$$
where, $$\mu =$$  refractive index of the material of the prism.
$$\eqalign{ & A = {\text{angle of prism}} = {60^ \circ } \cr & \therefore \mu = \frac{{\sin \left( {\frac{{{{60}^ \circ } + {{30}^ \circ }}}{2}} \right)}}{{\sin \frac{{{{60}^ \circ }}}{2}}} = \frac{{\sin {{45}^ \circ }}}{{\sin {{30}^ \circ }}} \cr & = \frac{{\frac{1}{{\sqrt 2 }}}}{{\frac{1}{2}}} = \frac{2}{{\sqrt 2 }} = \sqrt 2 \cr} $$

When an object is held in between two plane mirrors $${M_1}$$ and $${M_2}$$ inclined at $$\angle \theta ,$$  more than one image is formed on account of multiple reflections of light from the mirrors. In general, total number of images $$\left( n \right)$$  formed is given by
$$\eqalign{ & n = \frac{{{{360}^ \circ }}}{\theta } - 1 \cr & {\text{Here,}}\,\,\theta = {60^ \circ } \cr & \therefore n = \frac{{{{360}^ \circ }}}{{{{60}^ \circ }}} - 1 = 6 - 1 = 5 \cr} $$

Suppose a light ray enters at $$A$$ and refracted beam is $$AB.$$  At the lateral face, the angle of incidence is $$\theta .$$ For no refraction at this face, $$\theta > C$$
$$\sin \theta > \sin C$$

$$\eqalign{ & {\text{but}}\,\,\theta + r = {90^ \circ }\,\,\left( {{\text{According to geometry of figure}}} \right) \cr & \Rightarrow \theta = \left( {{{90}^ \circ } - r} \right) \cr & \therefore \sin \left( {{{90}^ \circ } - r} \right) > \sin C \cr & {\text{or}}\,\,\cos r > \sin C\,......\left( {\text{i}} \right) \cr} $$
Now from Snell’s law,
$$\eqalign{ & \mu = \frac{{\sin i}}{{\sin r}} \cr & \Rightarrow \sin r = \frac{{\sin i}}{\mu } \cr} $$
As from trigonometry,
$$\eqalign{ & \cos r = \sqrt {1 - {{\sin }^2}r} \cr & \therefore \cos r = \sqrt {\left( {1 - \frac{{{{\sin }^2}i}}{{{\mu ^2}}}} \right)} \cr} $$
$$\therefore $$ Eq. (i) gives,
$$\eqalign{ & \sqrt {1 - \frac{{{{\sin }^2}i}}{{{\mu ^2}}}} > \sin C \cr & \Rightarrow 1 - \frac{{{{\sin }^2}i}}{{{\mu ^2}}} > {\sin ^2}C \cr & {\text{Also}}\,\,\sin C = \frac{1}{\mu } \cr & \therefore 1 - \frac{{{{\sin }^2}i}}{{{\mu ^2}}} > \frac{1}{{{\mu ^2}}} \cr} $$
$$\eqalign{ & {\text{or}}\,\,1 > \frac{1}{{{\mu ^2}}} + \frac{{{{\sin }^2}i}}{{{\mu ^2}}} \cr & {\text{or}}\,\frac{1}{{{\mu ^2}}}\left( {{{\sin }^2}i + 1} \right) < 1 \cr & {\text{or}}\,\,{\mu ^2} > {\sin ^2}i + 1 \cr} $$
The maximum value of $$\sin i$$  is 1. So,
$$\eqalign{ & \therefore {\mu ^2} > 2 \cr & {\text{or}}\,\,\mu > \sqrt 2 \cr} $$

When $$\theta = {90^ \circ }$$
then, $$\frac{{360}}{\theta } = \frac{{360}}{{90}} = 4$$
is an even number. The number of images formed is given by
$$\eqalign{ & n = \frac{{360}}{\theta } - 1 \cr & = \frac{{360}}{{90}} - 1 \cr & = 4 - 1 \cr & = 3 \cr} $$

Let $$x$$ be the length of the shadow of the boy, when he is at a distance $$y$$ from the lamp.

From similar triangles, we have
$$\eqalign{ & \frac{h}{x} = \frac{{3h}}{{x + y}} \cr & {\text{or}}\,\,x = \frac{y}{2}. \cr} $$
Thus, $$\frac{{dx}}{{dt}} = \frac{{\frac{{dy}}{{dt}}}}{2}$$

Given,
Thickness of rectangular glass plate $$\left( t \right) = 6\,cm$$
Distance of the object $$\left( u \right) = 8\,cm$$
Distance of the image $$\left( v \right) = 12\,cm$$
Let $$y =$$ apparent position of the silvered surface. From property of mirror,
Image distance of object from the mirror = distance of image from the mirror
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\,y + 8 = 12 + 6 - y \cr & \therefore y = 5\,cm \cr & \therefore \mu = \frac{{{\text{real depth}}}}{{{\text{apparent depth}}}} \cr & = \frac{6}{5} = 1.2 \cr} $$

A real, inverted and enlarged image of the object is formed by the objective lens of a compound microscope.

Ray optics, uses the geometry of straight lines to account for the macroscopic phenomena like rectilinear propagation, reflection, refraction etc. Ray optics can be taken as a limiting case of wave optics. This is primarily because wavelength of light is small.

The ray is partly reflected and partly refracted
$$\angle MOB = 180 - 2\,\theta $$

But the angle between refracted and reflected rays is
$$\angle POB.$$  Clearly is $$\angle POB$$  is less than $$\angle MOB.$$

$$\frac{{\left| {{f_1}} \right|}}{{\left| {{f_2}} \right|}} = \frac{2}{3}$$
$${{f_1}}$$ : focal length of convex lens.
$$\eqalign{ & \frac{1}{f} = \frac{1}{{{f_1}}} - \frac{1}{{{f_2}}} \cr & \Rightarrow \,\,\frac{1}{{30}} = \frac{1}{{{f_1}}} - \frac{2}{{3{f_1}}} \cr & {f_1} = 10\,cm,{f_2} = - 15\,cm \cr} $$