1

### JEE Main 2017 (Online) 8th April Morning Slot

Let the refractive index of a denser medium with respect to a rarer medium be n12 and its critical angle be θC . At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 90o. Angle A is given by :
A
${1 \over {{{\cos }^{ - 1}}\left( {\sin {\theta _C}} \right)}}$
B
${1 \over {{{\tan }^{ - 1}}\left( {\sin {\theta _C}} \right)}}$
C
${\cos ^{ - 1}}\,\left( {\sin {\theta _C}} \right)$
D
${\tan ^{ - 1}}\,\left( {\sin {\theta _C}} \right)$

## Explanation

Refractive index,

n12 = ${{{n_D}} \over {{n_R}}}$ = ${1 \over {\sin {\theta _c}}}$

$\Rightarrow$ $\,\,\,$ sin$\theta$c = ${{{n_R}} \over {{n_D}}}$ . . . . . (1)

From Snell's law,

nD sinA = nR sin r

${{{n_R}} \over {{n_D}}}$ = ${{\sin A} \over {\sin \left( {{{90}^o} - A} \right)}}$    [as    r = 90o $-$ A]

$\Rightarrow $$\,\,\, {{{n_R}} \over {{n_D}}} = \tan A \therefore\,\,\, From (1) we get, tan A = sin \theta c \Rightarrow$$\,\,\,$ A = tan$-$1 (sin $\theta$c)
2

### JEE Main 2017 (Online) 9th April Morning Slot

A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000 $\mathop A\limits^ \circ$ and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is :
A
3 mm
B
9 mm
C
4.5 mm
D
1.5 mm
3

### JEE Main 2018 (Offline)

The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 $\mu$m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)
A
100 $\mu$m
B
25 $\mu$m
C
50 $\mu$m
D
75 $\mu$m

## Explanation

Given 2$\theta$ = 60o

$\Rightarrow$ $\theta$ = 30o

We know,

$a$ sin $\theta$ = n $\lambda$

for first minima n = 1,

$\therefore\,\,\,$ At first minima

$a$ sin = $\lambda$

$\Rightarrow $$\,\,\, 10-6 \times sin 30o = \lambda \Rightarrow$$\,\,\,$ $\lambda$ = ${{{{10}^{ - 6}}} \over 2}$ m

Now after making a new slit,

$\therefore\,\,\,$ Fringe width, $\beta$ = ${{\lambda D} \over d}$

given, D = 50 cm and $\beta$ = 1 cm.

$\therefore\,\,\,$ 1 $\times$ 10$-$2 = ${{0.5 \times {{10}^{ - 6}} \times 50 \times {{10}^{ - 2}}} \over d}$

$\Rightarrow $$\,\,\, d = 25 \times 10-6 m = 25 \mu m. 4 MCQ (Single Correct Answer) ### JEE Main 2018 (Offline) Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be I/2. Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be I/8. The angle between polarizer A and C is : A 60o B 30o C 45o D 0o ## Explanation As after B intensity of light does not drops, it means both A and B are alligned in single line means their plane of polarization is same. Let C makes an angle \theta with A then C will make \theta with B also, as both A and B are alligned in a single line. So, after C intensity is = {{\rm I} \over 2} cos2\theta , and , intensity after B = {{\rm I} \over 2} cos2\theta \times cos2\theta According to question, {{\rm I} \over 2} cos4\theta = {{\rm I} \over 8} \Rightarrow \,\,\, CO4\theta = {{\rm I} \over 4} \Rightarrow$$\,\,\,$ cos$\theta$ = $= {1 \over {\sqrt 2 }}$ = cos45o

$\therefore\,\,\,$ $\theta$ = 45o

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