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

$$\eqalign{ & {\mu _g}\sin i = {\mu _{air}}\sin {90^ \circ } \cr & {m_g} = \frac{1}{{\sin i}} \cr} $$

Power of combination is given by
$$\eqalign{ & P = {P_1} + {P_2} \cr & = \left( { - 15 + 5} \right)D \cr & = - 10\,D. \cr & {\text{Now, }}P = \frac{1}{f} \cr & \Rightarrow \,\,f = \frac{1}{P} \cr & = \frac{1}{{ - 10}}metre \cr & \therefore \,\,f = - \left( {\frac{1}{{10}} \times 100} \right)cm \cr & = - 10\,cm. \cr} $$

It is given that, $$\frac{{{I_2}}}{{{I_1}}} = n \Rightarrow {I_2} = n{I_1}$$
$$\therefore $$ Ratio of intensities is given by
$$\eqalign{ & \frac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}} = \frac{{{{\left( {\sqrt {{I_2}} + \sqrt {{I_1}} } \right)}^2} - {{\left( {\sqrt {{I_2}} - \sqrt {{I_1}} } \right)}^2}}}{{{{\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)}^2} + \left. {{{\left( {\sqrt {{I_2}} - \sqrt {{I_1}} } \right)}^2}} \right)}} \cr & = \frac{{{{\left( {\sqrt {\frac{{{I_2}}}{{{I_1}}}} + 1} \right)}^2} - {{\left( {\sqrt {\frac{{{I_2}}}{{{I_1}}}} - 1} \right)}^2}}}{{{{\left( {\sqrt {\frac{{{I_2}}}{{{I_1}}}} + 1} \right)}^2} + {{\left( {\sqrt {\frac{{{I_2}}}{{{I_1}}}} - 1} \right)}^2}}} \cr & = \frac{{{{\left( {\sqrt n + 1} \right)}^2} - {{\left( {\sqrt n - 1} \right)}^2}}}{{{{\left( {\sqrt n + 1} \right)}^2} + {{\left( {\sqrt n - 1} \right)}^2}}} \cr & = \frac{{2\sqrt n }}{{n + 1}} \cr} $$

Maximum number of reflection $$ = \frac{{2\sqrt 3 }}{x}$$
where $$x = 0.2\tan {30^ \circ } = \frac{{0.2}}{{\sqrt 3 }}.$$

KEY CONCEPT : The resolving power of a telescope $$R.P = \frac{D}{{1.22\lambda }}$$   where $$D$$ = diameter of the objective lens
$$\lambda $$ = wave length of light
Clearly, larger the aperture, larger is the value of $$D,$$ more is the resolving power or resolution.

$$\eqalign{ & {\text{As, we know that}}\,\,\mu = \frac{{\sin \left( {\frac{{A + {D_m}}}{2}} \right)}}{{\sin \frac{A}{2}}} \cr & \Rightarrow \cot \frac{A}{2} = \frac{{\sin \left( {\frac{{A + {D_m}}}{2}} \right)}}{{\sin \frac{A}{2}}} \cr & \Rightarrow \frac{{\cot \frac{A}{2}}}{{\sin \frac{A}{2}}} = \frac{{\sin \left( {\frac{{A + {D_m}}}{2}} \right)}}{{\sin \frac{A}{2}}} \cr & \Rightarrow \sin \left( {\frac{\pi }{2} - \frac{A}{2}} \right) = \sin \left( {\frac{{A + {D_m}}}{2}} \right) \cr & \Rightarrow \frac{\pi }{2} - \frac{A}{2} = \frac{A}{2} + \frac{{{D_m}}}{2} \cr & \Rightarrow {D_m} = \pi - 2A \cr & \therefore {D_m} = {180^ \circ } - 2A \cr} $$

Applying Snell’s law at $$Q$$
$$\eqalign{ & n = \frac{{\sin {{90}^ \circ }}}{{\sin C}} = \frac{1}{{\sin C}} \cr & \therefore \,\,\sin C = \frac{1}{n} = \frac{{\sqrt 3 }}{2} \cr & \therefore \,\,C = {60^ \circ } \cr} $$
Applying Snell’s Law at $$P$$
$$\eqalign{ & n = \frac{{\sin \theta }}{{\sin \left( {{{90}^ \circ } - C} \right)}} \cr & \Rightarrow \,\,\sin \theta = n \times \sin \left( {{{90}^ \circ } - C} \right);\,{\text{from }}\left( 1 \right) \cr & \Rightarrow \,\,\sin \theta = n\cos \cr & \therefore \,\,\theta = {\sin ^{ - 1}}\left[ {\frac{2}{{\sqrt 3 }} \times \cos {{60}^ \circ }} \right] \cr & {\text{or, }}\theta = {\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) \cr} $$

$$C = {\sin ^{ - 1}}\left( {\frac{1}{{_2^1\mu }}} \right)\,\,\,.....\left( {\text{i}} \right)$$
Applying Snell's law at $$P,$$ we get
$$\eqalign{ & _2^1\mu = \frac{{\sin r'}}{{\sin i}} \cr & = \frac{{\sin \left( {90 - r} \right)}}{{\sin r}}\left[ {\because \,\,i = r,r' + r = {{90}^ \circ }} \right] \cr & \because \,\,_2^1\mu = \frac{{\cos r}}{{\sin r}}\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii)
$$C = {\sin ^{ - 1}}\left( {\tan r} \right)$$

Here, total thickness $$= t$$
Refractive index $$ = \mu $$
Speed of light in glass plate $$ = \frac{c}{\mu }$$
$$\left[ {\because v = \frac{{{\text{speed of light in vacuum}}}}{{{\text{refractive index of medium}}}}} \right]$$
$$\therefore $$ Time taken by light to travel this thickness of glass
$$ = \frac{t}{{\left( {\frac{c}{\mu }} \right)}} = \frac{{\mu t}}{c}$$

When the ray enters a glass slab from air, its frequency remains unchanged.
Since glass slab in an optically denser medium, the velocity of light decreases and therefore we can conclude that the wavelength decreases.
$$\left( {\because \nu = \nu \lambda } \right)$$