Solution:

Ans. (i) The diagram, showing polarisation by reflection is as shown. [Here the reflected and refracted rays are at right angle to each other.

$$ \begin{array}{ll} \therefore & r=\left(\frac{\pi}{2}-i_{B}\right) \ \therefore & \mu=\left(\frac{\sin i_{B}}{\sin r}=\tan i_{B}\right) \end{array} $$

Thus light gets totally polarised by reflection when it is incident at an angle $i_{B}$ (Brewster’s angle), where $i_{B}=\tan ^{-1} \mu$

(ii) The angle of incidence, of the ray, on striking the face $\mathrm{AC}$ is $i=60^{\circ}$ (as from figure)

Also, relative refractive index of glass, with respect to the surrounding water, is

For total internal reflection, the required critical angle, in this case, is given by

$$ \sin i_{C}=\frac{1}{\mu}=\frac{8}{9} \simeq 0.89 $$

$\therefore \quad i<i_{C}$

Hence the ray would not suffer total internal reflection on striking the face $A C$.

[The student may just write the two conditions needed for total internal reflection without analysis of the given case.

The student may be awarded $(1 / 2+1 / 2)$ mark in such a case.

[CBSE Marking Scheme 2018]

AT Q. 2. (i) Monochromatic light of wavelength $589 \mathrm{~nm}$ is incident from air on a water surface. If $\mu$ for water is 1.33 , find the wavelength, frequency and speed of the refracted light.

(ii) A double convex lens is made of a glass of refractive index 1.55, with both faces of the same radius of curvature. Find the radius of curvature required, if the focal length is $20 \mathrm{~cm}$.

U [OD I, II, III, 2016]

Ans. (i)

$$ \lambda=\frac{589 \mathrm{~nm}}{1 \cdot 33}=442 \cdot 8 \mathrm{~nm} 1 / 2 $$

Frequency,

$$ v=\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{589 \mathrm{~nm}} $$

$$ =5.09 \times 10^{14} \mathrm{~Hz} $$

$1 / 2$

Speed

$$ v=\frac{3 \times 10^{8}}{1.33} \mathrm{~m} / \mathrm{s} $$

(ii)

$$ \begin{aligned} &=2 \cdot 25 \times 10^{8} \mathrm{~m} / \mathrm{s} \ & \frac{1}{f}=\left[\frac{\mu_{2}}{\mu_{1}}-1\right]\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \ & \frac{1}{20}=\left[\frac{1.55}{1}-1\right] \frac{2}{R} \ & R=(20 \times 1 \cdot 10) \mathrm{cm}=-22 \mathrm{~cm} \ & 1 / 2 \end{aligned} $$

$$ \therefore \quad \frac{1}{20}=\left[\frac{1.55}{1}-1\right] \frac{2}{R} $$

$$ \therefore \quad R=(20 \times 1 \cdot 10) \mathrm{cm}=-22 \mathrm{~cm} $$

[CBSE Marking Scheme 2016]