In Young's double-slit experiment,the two sli | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: Slit separation $d = 2 \, mm = 2 	imes 10^{-3} \, m$,Screen distance $D = 2 \, m$,$\lambda_1 = 12000 	imes 10^{-10} \, m$,$\lambda_2 = 10

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Given: Slit separation $d = 2 \, mm = 2 	imes 10^{-3} \, m$,Screen distance $D = 2 \, m$,$\lambda_1 = 12000 	imes 10^{-10} \, m$,$\lambda_2 = 10000 	imes 10^{-10} \, m$.For bright fringes to coincide,the position $x$ must satisfy: $x = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$.This implies $n_1 \lambda_1 = n_2 \lambda_2$,or $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}$.Substituting the values: $\frac{n_1}{n_2} = \frac{10000}{12000} = \frac{5}{6}$.For the minimum distance,we take the smallest integers $n_1 = 5$ and $n_2 = 6$.Now,calculate $x$: $x = \frac{n_1 \lambda_1 D}{d} = \frac{5 	imes 12000 	imes 10^{-10} 	imes 2}{2 	imes 10^{-3}}$.$x = 5 	imes 12000 	imes 10^{-7} \, m = 60000 	imes 10^{-7} \, m = 6 	imes 10^{-3} \, m = 6 \, mm$.

Similar Questions

In the Young's double slit experiment,for which colour is the fringe width the least?

Minimum thickness of a mica sheet having $\mu = 3/2$ which should be placed in front of one of the slits in $YDSE$ is required to reduce the intensity at the centre of the screen to half of the maximum intensity is-

In Young's double-slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $k$ units; $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\lambda/4$ will be:

In a Young's double-slit experiment,the distance between the two slits is $d = \lambda / 4$,where $\lambda$ is the wavelength of the light used. The initial phase difference is $\pi / 4$. What is the intensity at $\theta = 30^o$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module