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

$$\eqalign{ & m = - \frac{v}{u} = - \left( {\frac{f}{{u - f}}} \right) \cr & {\text{Now}}\,\,{m_1} = - \left( {\frac{f}{{25 - f}}} \right)\,.......\left( {\text{i}} \right) \cr & {\text{and}}\,\,{m_2} = - \left( {\frac{f}{{40 - f}}} \right)\,.......\left( {{\text{ii}}} \right) \cr & \therefore \frac{{{m_1}}}{{{m_2}}} = \frac{{40 - f}}{{25 - f}} \cr & {\text{or}}\,\,4 = \frac{{40 - f}}{{25 - f}} \cr & {\text{or}}\,\,f = 20\,cm. \cr} $$

Angle of incidence is given by
$$\eqalign{ & \cos \left( {\pi - i} \right) = \frac{{\left( {6\sqrt 3 \hat i + 8\sqrt 3 \hat j - 10\hat k} \right).\hat k}}{{20}} \cr & - \cos i = - \frac{1}{2} \cr & \angle \,i = {60^ \circ } \cr} $$
From Snell's law,
$$\eqalign{ & \sqrt 2 \sin i = \sqrt 3 \sin r \cr & \angle \,r = {45^ \circ } \cr} $$

$$\eqalign{ & {f_1} = + 40\,cm = 0.4\,m\,\,\left( {{\text{for}}\,{\text{convex}}\,{\text{lens}}} \right) \cr & {f_2} = - 25\,cm = - 0.25\,m\,\,\left( {{\text{for}}\,{\text{convcave}}\,{\text{lens}}} \right) \cr} $$

$$\therefore $$ Focal length $$\left( f \right)$$  of the combination
$$\eqalign{ & \frac{1}{f} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} = \frac{1}{{0.40}} - \frac{1}{{0.25}} = \frac{{0.25 - 0.4}}{{0.40 \times 0.25}} \cr & = - \frac{{0.15}}{{0.1}} = - 1.5\,{\text{dioptre}}{\text{.}} \cr & \Rightarrow P = \frac{1}{f} = - 1.5\,{\text{dioptre}} \cr} $$

(A) Interference is a phenomenon in which two waves of same frequency superpose to give resultant intensity different from sum of their separate intensity. So, it cannot exhibit particle nature of light.
(B) Diffraction is a phenomenon in which light bends at sharp ends of an obstacle or a hole. So it also can’t exhibit particle nature of light.
(C) Polarisation of light is a property owing to which a light ray after emerging through a crystal (a special kind like tourmaline called polaroid) have vibrations in a plane perpendicular to its direction of propagation. So, it also can’t explain particle nature of light.
(D) Photoelectric effect states that light travels in the form of bundles or packets of energy, called photons. This effect is explained on the basis of quantum nature of light. So, it clearly explains the particle nature of light.

Velocity of light waves in material is
$$\nu = \nu \lambda \,......\left( {\text{i}} \right)$$
Refractive index of material is
$$\mu = \frac{c}{\nu }\,......\left( {{\text{ii}}} \right)$$
where, $$c$$ is speed of light in vacuum or air.
$$\eqalign{ & {\text{or}}\,\,\mu = \frac{c}{{\nu \lambda }}\,......\left( {{\text{iii}}} \right) \cr & {\text{Given,}}\,\,\nu = 2 \times {10^{14}}Hz \cr & \lambda = 5000\,\mathop {\text{A}}\limits^ \circ = 5000 \times {10^{ - 10}}m, \cr & {\text{and}}\,\,c = 3 \times {10^8}m/s \cr} $$
Hence, from Eq. (iii), we get
$$\mu = \frac{{3 \times {{10}^8}}}{{2 \times {{10}^{14}} \times 5000 \times {{10}^{ - 10}}}} = 3.00$$

Refractive index,
$$\mu = \frac{c}{\nu } = \frac{{\nu {\lambda _\nu }}}{{\nu {\lambda _m}}}$$
As,
$$c =$$  velocity of light in vacuum
$$\nu = $$  velocity of light in medium
$$\nu = \nu \lambda $$  and $$\nu $$ remains constant during refraction
$$\eqalign{ & \therefore {\lambda _m} = \frac{{{\lambda _\nu }}}{\mu } \cr & \Rightarrow {\lambda _m} < {\lambda _\nu }\,\,\left( {\because \mu = 1,\,{\text{given}}} \right) \cr} $$
Hence, the wavelength decreases in second medium.

Distance of image from the plane surface is as follows :

$${x_1} = \frac{4}{{1.6}} = 2.5\,cm\left( {{d_{{\text{app}}}} = \frac{{{d_{{\text{actual}}}}}}{\mu }} \right)$$
For the curved side
$$\eqalign{ & \frac{{1.6}}{4} + \frac{1}{{{x^2}}} = \frac{{1 - 1.6}}{8} \cr & \therefore {x_2} \approx - 3.0\,cm \cr} $$
The minus sign means the image is on the object side.
$$\therefore {I_1}{I_2} = \left( {8 - 2.5 - 3.0} \right)cm = 2.5\,cm$$

NOTE : For minimum deviation, incident angle is equal to emerging angle and $$QR$$  is parallel to base.

Shift $$ = \left( {\ell - m} \right)\left( {1 - \frac{1}{{{n_1}}}} \right) + m\left( {1 - \frac{1}{{{n_2}}}} \right) = 0$$

Applying Snell's law at $$P,$$

$$\eqalign{ & \mu = \frac{{\sin r}}{{\sin {{30}^ \circ }}} \cr & \sin r = \frac{{1.44}}{2} = 0.72 \cr & \therefore \,\,\delta = r - {30^ \circ } \cr & = {\sin ^{ - 1}}\left( {0.72} \right) - {30^ \circ } \cr} $$
∴ The rays make an angle of
$$2\delta = 2\left[ {{{\sin }^{ - 1}}\left( {0.72} \right) - {{30}^ \circ }} \right]$$     with each other.