Question: Q. 10. A ray of light, incident on an equilateral glass prism $\left(\mu_{g}=\sqrt{3}\right)$ moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray.
A [Delhi I, II, III 2012]
From the diagram, $r=30^{\circ}$
Also
$n_{21}=\frac{\sin i}{\sin r}$
or,
$$ \sqrt{3}=\frac{\sin i}{\sin 30^{\circ}} $$
or, $\quad \sin i=\sqrt{3} \times \frac{1}{2}$
or, $\quad i=60^{\circ}$
[CBSE Marking Scheme 2012]
Short Answer Type Questions-II
(3 marks each)
Q. 1. (i) Show using a proper diagram how unpolarized light can be linearly polarised by reflection from a transparent glass surface.
(ii) The figure shows a ray of light falling normally on on the face $A B$ of an equilateral glass prism having refractive index $\frac{3}{2}$, placed in water of refractive index $\frac{4}{3}$. Will this ray suffer total internal reflection on striking the face $A C$ ? Justify your answer.
A [CBSE 2018, 2012]
Show Answer
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]
