JEE Main 31-Aug-2021 Shift 2 Solved Paper

Given, slit separation, d=0.3‌mm=0.3×10−3m
Distance between slit and screen, D = 1.5 m
Distance between 4th bright fringes =2Y4=2.4‌cm
Position of 4th fringe,
Y4=

2.4

2

=1.2‌cm=1.2×10−2m
Using equation, Yn=

nλD

d

For n = 4
1.2×10−2=

4×λ×1.5

0.3×10−3

∴ Wavelength, λ=6×10−7m
Using c = λf
⇒f=

c

λ

=

3×108

6×10−7

=5×1014‌Hz
Hence, the frequency of light is 5×1014‌Hz.
So, correct answer is 5.