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

For refractive index of a index,
$$\mu = \frac{1}{{\sin {i_c}}} = \frac{1}{{\sin {{45}^ \circ }}} = \sqrt 2 $$

$$\eqalign{ & {\text{As,}}\,\,{\mu _{{\text{red}}}} = 1.39,\,{\mu _{{\text{green}}}} = 1.44,\,{\mu _{{\text{blue}}}} = 1.47 \cr & \therefore \left( {{\mu _{{\text{red}}}} = 1.39} \right) < \mu ,{\mu _{{\text{green}}}} > \mu ,\,{\mu _{{\text{blue}}}} > \mu \cr} $$
Thus, only red colour do not suffer total internal reflection.

The image $$I'$$ for first refraction (i.e., when the ray comes out of liquid) is at a depth of
$$ = \frac{{33.25}}{{1.33}} = 25\,cm\,\,\left[ {\because {\text{Apparent depth}} = \frac{{{\text{Real depth}}}}{\mu }} \right]$$
Now, reflection will occur at concave mirror. For this $$I'$$ behaves as an object.
$$\therefore u = - \left( {15 + 25} \right) = - 40\,cm\,\,{\text{and}}\,\,v = - \left[ {15 + \frac{{25}}{{1.33}}} \right]$$
Where $$\frac{{25}}{{1.33}}$$  is the real depth of the image.
Using mirror formula we get
$$\eqalign{ & \frac{1}{f} = \frac{1}{v} + \frac{1}{u}, \cr & f = - 18.31\,cm \cr} $$

For a thin prism $$D = \left( {\mu - 1} \right)A$$
Since, $${\lambda _b} < {\lambda _r}$$
$$\eqalign{ & \Rightarrow \,\,{\mu _r} < {\mu _b} \cr & \Rightarrow \,\,{D_1} < {D_2} \cr} $$

See figure. The ray will come out from $$CD$$  if it suffers total internal reflection at surface $$AD,$$  i.e., it strikes the surface $$AD$$  at critical angle $$C$$ ( the limiting case).

Applying Snell's law at $$P$$
$${n_1}\sin C = {n_2}\,{\text{or }}\sin C = \frac{{{n_2}}}{{{n_1}}}$$
Applying Snell's law at $$Q$$
$$\eqalign{ & {n_2}\sin \alpha = {n_1}\cos C \cr & \Rightarrow \,\,\sin \alpha = \frac{{{n_1}}}{{{n_2}}}\cos \left\{ {{{\sin }^{ - 1}}\left( {\frac{{{n_2}}}{{{n_1}}}} \right)} \right\} \cr & {\text{or }}\alpha = {\sin ^{ - 1}}\left[ {\frac{{{n_1}}}{{{n_2}}}\cos \left\{ {{{\sin }^{ - 1}}\left( {\frac{{{n_2}}}{{{n_1}}}} \right)} \right\}} \right] \cr} $$

By Lens maker's formula for convex lens
$$\eqalign{ & \frac{1}{f} = \left( {\frac{\mu }{{{\mu _L}}} - 1} \right)\left( {\frac{2}{R}} \right) \cr & {\text{for, }}{\mu _{{L_1}}} = \frac{4}{3},{f_1} = 4R \cr & {\text{for, }}{\mu _{{L_2}}} = \frac{5}{3},{f_2} = - 5R \cr & \Rightarrow \,\,{f_2} = \left( - \right)ve \cr} $$

As the first minimum is observed at an angle of $${30^ \circ }$$ in a diffraction pattern due to a single slit of width $$a.$$
i.e. $$n = 1,\,\theta = {30^ \circ }$$
$$\because $$ According to Bragg’s law of diffraction,
$$\eqalign{ & a\sin \theta = n\lambda \Rightarrow a\sin {30^ \circ } = \left( 1 \right)\lambda \,\,\left( {n = 1} \right) \cr & \Rightarrow a = 2\lambda \,.......\left( {\text{i}} \right)\,\left\{ {\because \sin {{30}^ \circ } = \frac{1}{2}} \right\} \cr} $$
For 1^st secondary maxima
$$ \Rightarrow a\sin {\theta _1} = \frac{{3\lambda }}{2} \Rightarrow \sin {\theta _1} = \frac{{3\lambda }}{{2a}}\,\,........\left( {{\text{ii}}} \right)$$
Substitute value of a from Eq. (i) to Eq. (ii), we get
$$\sin {\theta _1} = \frac{{3\lambda }}{{4\lambda }} \Rightarrow \sin {\theta _1} = \frac{3}{4} \Rightarrow {\theta _1} = {\sin ^{ - 1}}\left( {\frac{3}{4}} \right)$$

According to question diagram is shown as

$$\tan \theta = \frac{{\frac{x}{2}}}{2}$$
For small $$\theta $$ and when $$\theta $$ is counted in rad, $$\tan \theta \approx \theta $$
$$\eqalign{ & {\text{So,}}\,\,\theta = \frac{{\frac{x}{2}}}{2} \cr & \Rightarrow \frac{\lambda }{a} = \frac{x}{4} \cr & \Rightarrow x = \frac{{4\lambda }}{a} = \frac{{4 \times 600 \times {{10}^{ - 9}}}}{{{{10}^{ - 3}}}} \cr & = 24 \times {10^{ - 4}}m \cr & = 2.4 \times {10^{ - 3}}m \cr & = 2.4\,mm \cr} $$

$$\eqalign{ & \delta = \left( {{{45}^ \circ } - {{30}^ \circ }} \right) + \left( {{{180}^ \circ } - {{60}^ \circ }} \right) + \left( {{{45}^ \circ } - {{30}^ \circ }} \right) \cr & = {150^ \circ }\,{\text{clockwise}}{\text{.}} \cr} $$

Use the concepts related to image formation by spherical mirrors.

The intermediate image in compound microscope is real, inverted and magnified.