1

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

A single slit of width b is illuminated by a coherent monochromatic light of wavelength $\lambda$. If the second and fourthminima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum ? (i.e. distance between first minimum on either side of the central maximum)
A
1.5 cm
B
3.0 cm
C
4.5 cm
D
6.0 cm

## Explanation

For single slit diffraction, sin$\theta$ = ${{n\lambda } \over b}$

From central maxima the position of nth minima = ${{n\lambda D} \over b}$

Now when,

n = 2, then x2 = ${{2\lambda D} \over b}$ = 0.03 . . . .(1)

n = 4, then x4 = ${{4\lambda D} \over b}$ = 0.06 . . . .(2)

Performing (2) $-$ (1) we get,

x4 $-$ x2 = ${{2\lambda D} \over b}$ = 0.03

$\therefore\,\,\,$ ${{\lambda D} \over b}$ = ${{0.03} \over 2}$

As, width of crntral maximum

= ${{2\lambda D} \over b}$

= 2 $\times$ ${{0.03} \over 2}$

= 0.03 m

= 3 cm
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