Yellow light of wavelength 590 nm passes through a diffraction grating and makes an interference pattern on a screen 80 cm away. The first bright fringes are 1.9 cm from the cen…

Question

asked Nov 9, 2021 45.2k views

1 Answer

Paulo Morgado · Answer 1 · 2021-11-13T17:40:53+0000

Answer:B

Step-by-step explanation:

Given

Wavelength of light
$\lambda =590\ nm$

Screen distance
$L=80\ cm$

First fringe is at a distance
$y_1=1.9\ cm$

No of lines per mm is given by N

N=(1)/(d)

where d=slit width

From N-slits Experiment

$\sin \theta _m=(m\lambda )/(d)$

$d=(m\lambda )/(\sin \theta _m)-----1$

Position of bright fringe is given by

$y=\tan \theta _m\cdot L$

$\tan \theta _m=(y)/(L)$

$\theta _m=\tan^(-1)((y)/(L))$

Put the value of
$\theta _m$ in eq. 1

$d=(m\lambda )/(\sin (\tan^(-1)((y)/(L))))$

Therefore
N=d^(-1)

$N=(\sin (\tan^(-1)((y)/(L))))/(m\lambda )$

for
m=1

$N=(\sin (\tan^(-1)((1.9* 10^(-2))/(0.8))))/(1* 590* 10^(-9))$

$N=40243\ line/m$

$N=40\ line/mm$