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

The lateral displacement is given by
$$d = \frac{t}{{\cos r}}\sin \left( {i - r} \right)$$
For small angle $$i,$$ angle $$r$$ is also small, and so
$$d = t\left( {i - r} \right) = t\,i\left( {1 - \frac{r}{i}} \right)$$

For total internal reflection, $$i > c$$   (initial angle)
$$\eqalign{ & {\text{So,}}\,\,\sin i > \sin c \cr & \Rightarrow \sin {45^ \circ } > \frac{1}{\mu } \cr & \Rightarrow \mu > \sqrt 2 \cr & \Rightarrow \mu > 1.4 \cr} $$

In figure, $$O$$ is the point source of light, $$K$$ is centre of disc of radius $$r,$$ $$OK = h.$$   The source will not be seen at all when the height is such that $$\angle KOL = C.$$   (i.e. critical angle between water and air.)

From figure,
$$\sin C = \frac{{KL}}{{OL}} = \frac{r}{{\sqrt {{r^2} + {h^2}} }}$$
For total internal reflection of light,
$$\eqalign{ & \sin C = \frac{1}{\mu } = \frac{r}{{\sqrt {{r^2} + {h^2}} }} \cr & \therefore {r^2} + {h^2} = {\mu ^2}{r^2} \cr & {\text{or}}\,\,{h^2} = {r^2}\left( {{\mu ^2} - 1} \right) \cr & {\text{Given,}}\,\,h = 4\;m,\mu = \frac{5}{3} \cr & {\text{or}}\,\,{r^2} = \frac{{{h^2}}}{{{\mu ^2} - 1}} = \frac{{{{\left( 4 \right)}^2}}}{{{{\left( {\frac{5}{3}} \right)}^2} - 1}} \cr & {\text{So,}}\,\,{r^2} = 9 \cr & \therefore r = 3\;m \cr & \therefore {\text{Diameter}}\,{\text{of}}\,{\text{disc}} = 2r = 6\;m \cr} $$

Given $$d = 1\,mm = 1 \times {10^{ - 3}}m$$
$$D = 1\,m,\lambda = 500\,nm = 5 \times {10^{ - 7}}m$$
As width of central maxima = width of 10 maxima
$$\eqalign{ & \therefore \frac{{2D\lambda }}{a} = 10\left( {\frac{{\lambda D}}{d}} \right) \cr & \Rightarrow a = \frac{d}{5} = \frac{{{{10}^{ - 3}}}}{5} = 0.2 \times {10^{ - 3}}m \cr & a = 0.2\,mm \cr} $$

Dispersive power $$\left( \omega \right) = \frac{{{n_v} - {n_r}}}{{{n_y} - 1}} = \frac{{1.521 - 1.510}}{{1.515 - 1}} = 0.02$$

Min. width of plane mirror to see full face
$$ = \frac{{D - d}}{2} = \frac{{10 - 4}}{2} = 3\,cm.$$

$$\sin q = \frac{{0.25}}{{25}} = \frac{1}{{100}}$$

Resolving power $$ = \frac{{1.221}}{{2\,m\sin q}} = 30\,\mu m.$$

From lens maker's formula
$$\frac{1}{f} = \left( {{\mu _g} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\,.......\left( {\text{i}} \right)$$
When convex lens is dipped in a liquid of refractive index $$\left( {{\mu _l}} \right),$$  then its focal length becomes
$$\frac{1}{{{f_l}}} = \left( {\frac{{{\mu _g}}}{{{\mu _l}}} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)$$
$${\text{or}}\,\,\frac{1}{{{f_l}}} = \frac{{\left( {{\mu _g} - {\mu _l}} \right)}}{{{\mu _l}}}\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (i) by Eq. (ii), we get
$$\frac{{{f_l}}}{f} = \frac{{\left( {{\mu _g} - 1} \right){\mu _l}}}{{\left( {{\mu _g} - {\mu _l}} \right)}}\,......\left( {{\text{iii}}} \right)$$
But it is given that refractive index of lens is equal to refractive index of liquid i.e. $${\mu _g} = {\mu _l}.$$
Hence, Eq. (iii) gives,
$$\frac{{{f_l}}}{f} = \frac{{\left( {{\mu _g} - 1} \right){\mu _l}}}{0} = \infty \,\,\left( {{\text{infinity}}} \right)$$

Given, $${\mu _g} = 1.5$$

$${\mu _{{\text{oil}}}} = 1.7 \Rightarrow R = 20\,cm$$
From Lens Maker's formula for the plano convex lens
$$\eqalign{ & \frac{1}{f} = \left( {\mu - 1} \right)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right] \cr & {\text{Here,}}\,\,{R_1} = R \cr} $$
and for plane surface $${R_2} = \infty $$
$$\therefore \frac{1}{{{f_{{\text{lens}}}}}} = \left( {1.5 - 1} \right)\left( {\frac{1}{R} - 0} \right) \Rightarrow \frac{1}{{{f_{{\text{lens}}}}}} = \frac{{0.5}}{R}$$
When the intervening medium is filled with oil, then focal length of the concave lens formed by the oil
$$\eqalign{ & \frac{1}{{{f_{{\text{concave}}}}}} = \left( {1.7 - 1} \right)\left( { - \frac{1}{R} - \frac{1}{R}} \right) \cr & = - 0.7 \times \frac{2}{R} = \frac{{ - 1.4}}{R} \cr} $$
Here, we have two concave surfaces
$$\eqalign{ & {\text{So,}}\,\,\frac{1}{{{f_{{\text{eq}}}}}} = 2 \times \frac{1}{f} + \frac{1}{f} \cr & = 2 \times \frac{{0.5}}{R} + \left( {\frac{{ - 1.4}}{R}} \right) = \frac{1}{R} - \frac{{1.4}}{R} = - \frac{{0.4}}{R} \cr & \therefore {f_{{\text{eq}}}} = - \frac{R}{{0.4}} = - \frac{{20}}{{0.4}} = - 50\,cm \cr} $$

$$C = {\sin ^{ - 1}}\left( {\frac{1}{{_2^1\mu }}} \right)\,\,\,.....\left( {\text{i}} \right)$$
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)
$$C = {\sin ^{ - 1}}\left( {\tan r} \right)$$