In a Young's double slit experiment,a student | vedclass

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

Question

(C) The width of the segment on the screen is given by $y = n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Thus,$y = n \lambda \left

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) The width of the segment on the screen is given by $y = n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Thus,$y = n \lambda \left(\frac{D}{d}ight)$.Since the segment length $y$ and the experimental setup parameters $D$ and $d$ remain constant,we have $n_1 \lambda_1 = n_2 \lambda_2$.Given $n_1 = 8$,$\lambda_1 = 600 \ nm$,and $\lambda_2 = 400 \ nm$.Substituting the values: $8 	imes 600 = n_2 	imes 400$.$n_2 = \frac{8 	imes 600}{400} = \frac{4800}{400} = 12$.Therefore,the student will observe $12$ fringes.

In a Young's double slit experiment,a student observes $8$ fringes in a certain segment of the screen when a monochromatic light of $600 \ nm$ wavelength is used. If the wavelength of light is changed to $400 \ nm$,then the number of fringes he would observe in the same region of the screen is:

Similar Questions

In Young's double-slit experiment,if the separation between the slit and the screen increases,the fringe width:

The distance between two coherent sources is $1 \ mm$. The screen is placed at a distance of $1 \ m$ from the sources. If the distance of the third bright fringe is $1.2 \ mm$ from the central fringe, the wavelength of light used is:

Obtain the formula for the path difference at a point on the screen in Young's double-slit experiment in terms of $x$,$d$,and $D$.

An interference pattern was made by using red light. If the red light is replaced with blue light,the fringes will become

Two slits are made $1\, mm$ apart and the screen is placed $1\, m$ away. The fringe separation when monochromatic light of wavelength $500\, nm$ is used is

Vedclass Test Series

Exam Paper Generator

Online Exam Module