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

Here, $${f_0} = 2\,cm{\text{ and}}\,{\text{ }}{f_e} = 3\,cm.$$
Using lens formula for eye piece
$$\eqalign{ & \Rightarrow \,\,\frac{{ - 1}}{{{u_e}}} + \frac{1}{\propto } = \frac{1}{3} \cr & \Rightarrow \,\,{u_e} = - 3\,cm \cr} $$
But the distance between objective and eye piece is $$15\,cm$$  (given)
∴ Distance of image formed by the objective
$$= v = 15 - 3 = 12\,cm.$$
Let $$u$$ be the object distance from objective, then for objective lens
$$\eqalign{ & - \frac{1}{{{u_0}}} + \frac{1}{{{v_0}}} = \frac{1}{{{f_0}}}\,\,{\text{or }}\frac{{ - 1}}{u} + \frac{1}{{12}} = \frac{1}{2} \cr & \Rightarrow \,\,\frac{{ - 1}}{u} = \frac{1}{2} - \frac{1}{{12}} \cr & = \frac{5}{{12}} \cr & u = - \frac{{12}}{5} \cr & = - 2.4\,cm \cr} $$

Applying Snell's law at $$A$$
$$\eqalign{ & 1 \times \sin {45^ \circ } = \sqrt 2 \times \sin {r_1} \cr & \therefore \,\,{r_1} = {30^ \circ }\,\,\,\,.....\left( {\text{i}} \right) \cr} $$

Applying Snell's law at $$B$$
$$\eqalign{ & \sqrt 2 \sin C = 1 \times \sin {90^ \circ } \cr & \therefore \,\,C = {45^ \circ }\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr & {\text{In }}\Delta \,AMB,{90^ \circ } + \theta + {r_1} + \left( {{{90}^ \circ } - C} \right) = {180^ \circ }\left( {{\text{From fig}}{\text{.}}} \right) \cr & \therefore \,\,\theta = {15^ \circ } \cr} $$

According to question,
5^th dark fringe in air = 8 bright fringe in the medium
$$\eqalign{ & \left( {2 \times 5 - 1} \right)\frac{{\lambda D}}{{2d}} = 8\frac{{\lambda D}}{{\mu d}} \Rightarrow 9\frac{{\lambda D}}{{2d}} = 8\frac{{\lambda D}}{{\mu d}} \cr & \Rightarrow \frac{9}{2} = \frac{8}{\mu } \Rightarrow \mu = \frac{{8 \times 2}}{9} \cr} $$
∴ Refractive index of the medium,
$$\mu = \frac{{16}}{9} = 1.7777 \approx 1.78$$

Polarisation phenomenon is not common to sound and light. As light waves are transverse waves, so phenomenon of polarisation occurs only in light.

$$\eqalign{ & {{\vec v}_{{\text{obj,}}\,{\text{mirror}}}} = 4\hat i + 9\hat j\frac{{dx}}{{dt}} = 4,\frac{{dy}}{{dt}} = 9;u = - x \cr & - \frac{1}{{10}} = \frac{1}{V} + \frac{1}{{ - x}};V = \frac{{ - 10x}}{{x - 10}} \cr & {v_{1x}} = \frac{{dV}}{{dt}} = {\left( {\frac{{10}}{{x - 10}}} \right)^2}\left( {\frac{{dx}}{{dt}}} \right);{v_{1x}} = - 16 \cr & m = - \frac{V}{{ - x}} = \frac{{ - 10}}{{x - 10}} = \frac{{y_1^0}}{y} \cr & {v_{1y}} = - \left( {\frac{{10}}{{x - 10}}} \right)\frac{{dy}}{{dt}};{v_{1y}} = - 18 \cr & {{\vec v}_{{\text{image, mirror}}}} = - 16\hat i - 18\hat j; \cr & {{\vec v}_{{\text{image}}}} = - 12\hat i - 16\hat j \cr & |v| = 20\,cm/s \cr} $$

The rays coming from the point object fall on the glass-air interface normally and hence pass undeviated. Therefore if we retrace the path of the refracted rays backwards, the image will be formed at the centre only.

$$\eqalign{ & \frac{{{\mu _2}}}{V} - \frac{{{\mu _1}}}{{ - {u_0}}} = \frac{{{\mu _2} - {\mu _1}}}{{ - 2R}} \cr & \frac{{{\mu _3}}}{{{V_f}}} - \frac{{{\mu _2}}}{V} = \frac{{{\mu _3} - {\mu _2}}}{{ - R}} \cr & \frac{{{\mu _3}}}{{{V_f}}} + \frac{{{\mu _1}}}{{{u_0}}} = \frac{{{\mu _1} - {\mu _2}}}{{2R}} + \frac{{{\mu _2} - {\mu _3}}}{R} \cr & \therefore \frac{{{\mu _1} - {\mu _2}}}{2} < {\mu _3} - {\mu _2} \cr & \Rightarrow {\mu _1} - {\mu _2} < 2{\mu _3} - 2{\mu _2} \cr & \Rightarrow {\mu _1} + {\mu _2} < 2{\mu _3} \cr} $$

Given, power $$\left( {{P_1}} \right) = 40\,D$$    and power $$\left( {{P_2}} \right) = 20\,D$$
We have $${P_{{\text{eq}}}} = {P_1} + {P_2}$$
$$ = 40 + 20 = 60\,D$$
$$\eqalign{ & {\text{So,}}\,\,\frac{1}{{{f_{{\text{eq}}}}}} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} \cr & \Rightarrow {f_{{\text{eq}}}} = \frac{{100}}{{{P_{{\text{eq}}}}}} = \frac{{100}}{{60}} = 1.67\,cm \cr} $$

A hypermetropic person or a long sighted person can see clearly only those objects which are at long distances, as if near point of his eye has shifted to some distant point. This defect arises on account of contraction of eye ball or increase in focal length of eye lens. To correct this defect, the person has to wear specs using a convex lens of suitable focal length given by
$$\frac{1}{f} = \frac{1}{D} - \frac{1}{d}$$
where $$d =$$  actual distance of distinct vision of defective eye, and
$$D =$$  distance of distinct vision of normal eye.
NOTE
A myopic person can see clearly only those objects which are at short distances from the eye. This is as if far point of the eye has shifted to some nearby point.
Oldage vision defect is also called astigmatism. It arises because curvature of the cornea plus eye lens refracting system is not the same in different planes. This defect is removed by using a cylindrical lens with its axis along the vertical.

From similar triangles $$APB$$  and $$CBD,$$  point $$P$$ will be just visible if $$CD = 4\,R$$
$$\eqalign{ & \mu \sin i = 1 \times \sin r \cr & \mu \sin \left( {\angle CDQ} \right) = 1 \times \sin \left( {\angle CDB} \right) \cr & \mu \times \frac{R}{{R\sqrt {17} }} = 1 \times \frac{{2R}}{{R\sqrt {20} }} \cr & \Rightarrow \mu = \frac{{2\sqrt {17} }}{{\sqrt {20} }} = \frac{{\sqrt {85} }}{5} \cr} $$