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

Spherical aberration occurs due to the inability of a lens to converge marginal rays of the same wavelength to the focus as it converges the paraxial rays. This can be done by using a circular annular mask over the lens.

$$\eqalign{ & + 5 = - \frac{v}{u} \cr & \Rightarrow \,\,v = - 5\,u \cr} $$
Using $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
$$\eqalign{ & \Rightarrow \,\,\frac{1}{{ - 5u}} + \frac{1}{u} = \frac{{ - 1}}{{0.4}} \cr & \therefore \,\,u = - 0.32\,m. \cr} $$

For convex mirror :

$$\frac{1}{v} + \frac{1}{{ - 30}} = \frac{1}{{ + 15}}\,{\text{or}}\,v = 10\,m$$
For concave mirror : $$\frac{1}{v} + \frac{1}{{ - 70}} = \frac{1}{{ - 15}}$$
$${\text{or}}\,v = - 19.09\,cm$$

Interference phenomenon is shown by both light and sound waves.

The refractive index of material of prism (from Snell's law) is $$\mu = \frac{{\sin i}}{{\sin r}}$$
Here, $$i = \frac{{A + {\delta _m}}}{2}\,\,{\text{and}}\,\,r = \frac{A}{2}$$
where, $$A$$ is the angle of prism and $${{\delta _m}}$$ the angle of minimum deviation.
In case of minimum deviation, refractive index of prism is given by
$$\eqalign{ & \mu = \frac{{\sin \left( {\frac{{A + {\delta _m}}}{2}} \right)}}{{\sin \frac{A}{2}}} \cr & {\text{Given,}}\,\mu = \sqrt 3 ,\,A = {60^ \circ }\,\,\left( {{\text{for}}\,{\text{prism}}} \right) \cr & {\text{Thus,}}\,\,\sqrt 3 = \frac{{\sin \left( {\frac{{60 + {\delta _m}}}{2}} \right)}}{{\sin {{30}^ \circ }}} \cr & {\text{or}}\,\,\sin \left( {\frac{{60 + {\delta _m}}}{2}} \right) = \frac{1}{2} \times \sqrt 3 \cr & {\text{or}}\,\,\sin \left( {\frac{{60 + {\delta _m}}}{2}} \right) = \sin {60^ \circ } \cr & {\text{or}}\,\,\frac{{60 + {\delta _m}}}{2} = 60 \cr & {\text{or}}\,\,{\delta _m} = 2 \times 60 - 60 = {60^ \circ } \cr} $$

According to Lens maker's formula, focal length of lens is given by
$$\frac{1}{f} = \left( {\mu - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\,......\left( {{\text{ii}}} \right)$$
We know that for planoconvex lens, the radius of curvature of plane surface is infinite,
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\,{R_2} = \infty . \cr & {\text{Given,}}\,\,{R_1} = 60\,cm,\mu = 1.6 \cr} $$
Substituting the given values in Eq. (i), we get
$$\frac{1}{f} = \left( {1.6 - 1} \right)\left( {\frac{1}{{60}} - \frac{1}{\infty }} \right) = 0.6 \times \frac{1}{{60}}$$
∴ Focal length of plane convex lens
$$f = \frac{{60}}{{0.6}} = 100\,cm$$

The ray $$OR$$  will retrace its path, when

$$\eqalign{ & \angle ARQ = {90^ \circ },\angle r = {30^ \circ } \cr & \therefore \angle AQR = {90^ \circ } - {30^ \circ } = {60^ \circ } \cr} $$
As from Snell's law,
$$\eqalign{ & \sin i \times 1 = \sin r \times \mu \cr & \sin i = \mu \sin r = \sqrt 2 \sin {30^ \circ }\,\left[ {{\text{Refractive index of air}} = 1} \right] \cr & = \sqrt 2 \times \frac{1}{2} = \frac{1}{{\sqrt 2 }} \cr & \therefore \angle i = {45^ \circ } \cr} $$

When electromagnetic wave enters in other medium, frequency remains unchanged while wavelength and velocity become $$\frac{1}{\mu }$$ times.
So, after entering from air to glass slab of refractive index $$\left( \mu \right),$$ frequency remains $$\nu ,$$ wavelength $$\lambda ' = \frac{\lambda }{\mu }$$   and velocity of light $$v' = \frac{v}{\mu }.$$

The focal length $$\left( {{f_1}} \right)$$  of the lens with $$n = 1.5$$   is given by
$$\eqalign{ & \frac{1}{{{f_1}}} = \left( {{n_1} - 1} \right)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right] \cr & = \left( {1.5 - 1} \right)\left[ {\frac{1}{{14}} - \frac{1}{\infty }} \right] \cr & = \frac{1}{{28}} \cr} $$
The focal length $$\left( {{f_2}} \right)$$  of the lens with $$n = 1.2$$   is given by
$$\eqalign{ & \frac{1}{{{f_2}}} = \left( {{n_2} - 1} \right)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right] \cr & = \left( {1.2 - 1} \right)\left[ {\frac{1}{\infty } - \frac{1}{{ - 14}}} \right] \cr & = \frac{1}{{70}} \cr} $$
The focal length $$F$$ of the combination is
$$\frac{1}{F} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} = \frac{1}{{20}}$$
Applying lens formula for the combination of lens
$$\eqalign{ & \frac{1}{V} - \frac{1}{U} = \frac{1}{F} \cr & \Rightarrow \,\,\frac{1}{V} - \frac{1}{{ - 40}} = \frac{1}{{20}} \cr & \Rightarrow \,\,V = 40\,cm \cr} $$

When the light is incident on glass - an interface at an angle less than critical angle a small part of light will be reflected and most part will be transmitted.
When the light is incident greater than the critical angle, it gets completed reflected (total internal reflection) These characteristics are depicted in option (C).