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

For microscope, $$m = \frac{L}{{{f_o}}}\frac{D}{{{f_e}}} \Rightarrow m \propto \frac{1}{{{f_o}}}$$
For telescope, $$m = \frac{{{f_o}}}{{{f_e}}}$$
$$m \propto {f_o}$$
The magnifying power of microscope will decrease but the magnifying power of telescope will increase.

For the image of point $$P$$ to be seen by the observer, it should be formed at point $$Q.$$

In $$\Delta QNS,\,NS = QS = 2h$$
$$\eqalign{ & \therefore \angle NQS = {45^ \circ } \cr & \therefore r = {45^ \circ } \cr} $$
Now in $$\Delta QMA,$$
$$\eqalign{ & \angle MQA = {45^ \circ } \cr & \therefore MA = QA = h \cr} $$
$$_2^1\mu = \frac{{\sin r}}{{\sin i}} = \frac{{\sin {{45}^ \circ }}}{{\sin i}}\,.......\left( {\text{i}} \right)$$
$$\eqalign{ & {\text{In}}\,\Delta PMB, \cr & P{M^2} = 4{h^2} + {h^2} = 5{h^2} \cr & \therefore \sin i = \frac{h}{{\sqrt 5 h}} = \frac{1}{{\sqrt 5 }}\,......\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii)
$$_2^1\mu = \sqrt {\frac{5}{2}} $$

For minimum deviation the ray in the prism is parallel to the base of the prism. This condition does not depend on the colour (or wave length) of incident radiation. So in both the cases, by geometry, $$r = {30^ \circ }.$$  So (A) is correct option.

Given, YDSE experiment, having two slits of width are in the ratio of $$1:25.$$
So, ratio of intensity,
$$\frac{{{I_1}}}{{{I_2}}} = \frac{{{W_1}}}{{{W_2}}} = \frac{1}{{25}} \Rightarrow \frac{{{I_2}}}{{{I_1}}} = \frac{{25}}{1}$$
$$\eqalign{ & \therefore \frac{{{I_{\max }}}}{{{I_{\min }}}} = \frac{{{{\left( {\sqrt {{I_2}} + \sqrt {{I_1}} } \right)}^2}}}{{{{\left( {\sqrt {{I_2}} - \sqrt {{I_1}} } \right)}^2}}} = {\left[ {\frac{{\sqrt {\frac{{{I_2}}}{{{I_1}}}} + 1}}{{\sqrt {\frac{{{I_2}}}{{{I_1}}}} - 1}}} \right]^2} \cr & \Rightarrow {\left[ {\frac{{5 + 1}}{{5 - 1}}} \right]^2} = {\left( {\frac{6}{4}} \right)^2} = \frac{{36}}{{16}} = \frac{9}{4} \cr & {\text{Thus,}}\,\,\frac{{{I_{\max }}}}{{{I_{\min }}}} = \frac{9}{4} \cr} $$

For the condition of maxima

$$\sin \theta = \frac{\lambda }{a}$$
From the geometry, $$\sin \theta = \theta = \frac{Y}{D}$$     (for small angle)
$$\eqalign{ & {\text{So,}}\,\,\frac{Y}{D} = \frac{\lambda }{a} \cr & \Rightarrow Y = \frac{{\lambda D}}{a} \cr} $$
Hence, width of central maxima $$ = 2Y = \frac{{2\lambda D}}{a}$$

The incident and emergent ray of a glass slab are parallel therefore, the angle remains the same.

KEY CONCEPT : $$\sin C = \frac{1}{\mu }\,{\text{and }}\mu \,\propto \,\frac{1}{\lambda }$$
$$\therefore \,\,\sin C\, \propto \,\lambda $$
For higher value of $$\lambda $$ the angle $$C$$ also increases

A telescope magnifies by making the object appearing closer.

Given, $${\lambda _1} = 12000\,\mathop {\text{A}}\limits^ \circ \,\,{\text{and}}\,\,{\lambda _2} = 10000\,\mathop {\text{A}}\limits^ \circ ,$$
$$D = 2\,m\,\,{\text{and}}\,\,d = 2\,mm = 2 \times {10^{ - 3}}m.$$
Fringe-width of first wavelength, $${\beta _1} = \frac{{{\lambda _1}D}}{d}$$
$$\eqalign{ & = \frac{{12000 \times {{10}^{ - 10}} \times 2}}{{2 \times {{10}^{ - 3}}}} \cr & = 1.2\,mm \cr} $$
Fringe-width of second wavelength, $$\beta = \frac{{{\lambda _2}D}}{d}$$
$$\eqalign{ & = \frac{{10000 \times {{10}^{ - 10}} \times 2}}{{2 \times {{10}^{ - 3}}}} \cr & = 1\,mm \cr} $$
At $$6\,mm$$  distance from central bright fringe, $${5^{th}}$$ fringe of first wavelength coincide with $${6^{th}}$$ fringe of second wavelength.

From Fig. (i) and (ii), its is clear that if the mirror moves distance $$'A',$$ then the image moves a distance $$'2A'.$$