Ray Optics MCQ Questions & Answers in Optics and Wave | Physics

For refraction at parallel interfaces
$${n_0}\sin \theta = \frac{{{n_0}}}{2}\sin \alpha = \frac{{{n_0}}}{6}\sin \beta = \frac{{{n_0}}}{8}\sin {90^ \circ }$$

$$\therefore \,\,\sin \theta = \frac{1}{8}$$

So, required distance $$ = d + d + d = 3d.$$

For convex lens,
$$\frac{1}{{30}} = \frac{1}{v} + \frac{1}{{60}} \Rightarrow \frac{1}{{60}} = \frac{1}{v}$$
Similarly for concave lens
$$\frac{1}{{ - 120}} = \frac{1}{v} - \frac{1}{{40}} \Rightarrow \frac{1}{v} = \frac{1}{{60}}$$
Virtual object $$10\,cm$$  behind plane mirror.
Hence real image $$10\,cm$$  infront of mirror or, $$60\,cm$$  from convex lens.

For an astronomical telescope,
where magnification, $$m = \frac{{{f_o}}}{{{f_e}}}$$
$${{f_o}} = $$  focal length of objective lens
$${{f_e}} = $$  focal length of eye piece lens
Length of telescope tube $$L = {f_o} + {f_e}$$
Given, $$m = 10,L = 44\,cm$$
$$\eqalign{ & \therefore \frac{{{f_o}}}{{{f_e}}} = 10\,\,{\text{and}}\,\,{f_o} + {f_e} = 44 \cr & \Rightarrow {f_e} = \frac{{{f_o}}}{{10}} \cr & {\text{So,}}\,\,{f_o} + \frac{{{f_o}}}{{10}} = 44 \cr & {\text{or}}\,\,\frac{{11{f_o}}}{{10}} = 44\,\,{\text{or}}\,\,{f_o} = 40\,cm \cr} $$

From the figure in $$\Delta \,ABC,\tan \beta = \frac{{AB}}{{AC}}$$
$$ \Rightarrow \,\,AB = AC\tan \beta ,2r = f\tan \beta $$
⇒ Area of image $$ = \pi {r^2} \propto {f^2}$$

There will be no refraction from $$P$$ to $$Q$$ and then from $$Q$$ to $$R$$ (all being identical). Hence the ray will now have the same deviation.

Applying Snell’s law at point $$P$$
$$\eqalign{ & {n_1}\sin \theta = {n_2}\sin {90^ \circ } \cr & \therefore \,\,{n_2} = 2.72 \times \frac{{\frac{{11.54}}{2}}}{{\sqrt {{{\left( {10} \right)}^2} + {{\left( {\frac{{11.54}}{2}} \right)}^2}} }} \cr & \therefore \,\,{n_2} = 1.36 \cr} $$

Applying Snell's law at $$P,$$

$$\eqalign{ & \mu = \frac{{\sin r}}{{\sin {{30}^ \circ }}} \cr & \sin r = \frac{{1.44}}{2} = 0.72 \cr & \therefore \delta = r - {30^ \circ } = {\sin ^{ - 1}}\left( {0.72} \right) - {30^ \circ } \cr} $$
$$\therefore $$ The rays make an angle of
$$2\delta = 2\left[ {{{\sin }^{ - 1}}\left( {0.72} \right) - {{30}^ \circ }} \right]\,\,{\text{with}}\,{\text{each}}\,{\text{other}}{\text{.}}$$

Angular limit of resolution of eye $$ = \frac{{{\text{wavelength of light}}}}{{{\text{diameter of eye lens}}}}$$
$${\text{i}}{\text{.e}}\,\,\theta = \frac{\lambda }{d}\,......\left( {\text{i}} \right)$$
If $$y$$ is the minimum distance between two points at distance $$D$$ from eye, then
Angular limit of resolution of eye
$$\theta = \frac{y}{d}\,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we have
$$\eqalign{ & \frac{y}{d} = \frac{\lambda }{d} \cr & {\text{or}}\,\,y = \frac{{\lambda D}}{d}\,......\left( {{\text{iii}}} \right) \cr} $$
Given, $$\lambda = 5000\,\mathop {\text{A}}\limits^ \circ = 5 \times {10^{ - 7}}m,D = 50\,m,$$
$$d = 2\,mm = 2 \times {10^{ - 3}}m$$
Substituting in Eq. (iii), we get
$$\eqalign{ & y = \frac{{5 \times {{10}^{ - 7}} \times 50}}{{2 \times {{10}^{ - 3}}}} \cr & = 12.5 \times {10^{ - 3}}m \cr & = 1.25\,cm \cr} $$

$$\eqalign{ & {{\vec V}_{{\text{im}}}} = \frac{{{V^2}}}{{{u^2}}}{{\vec V}_{{\text{om}}}} \cr & 0 - \frac{{40}}{{13}} = - \left( {\frac{{ - 10}}{{ - 10 + x}}} \right)\left( {10 - \frac{{40}}{{13}}} \right) \cr & \frac{{ - 10}}{{ - 10 + x}} = \pm \frac{2}{3} \cr & x = 25, - 5cm \cr} $$
Object is infront of mirror
Hence $$25\,cm.$$