In Young's double slit experiment,if the sepa | vedclass

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

Question

(D) The formula for fringe width in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is

Vedclass · Accepted Answer

four times

Vedclass · Answer

(D) The formula for fringe width in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance of the screen from the sources,and $d$ is the separation between the sources.Let the initial fringe width be $\beta = \frac{\lambda D}{d}$.According to the problem,the new distance $D^{\prime} = 2D$ and the new separation $d^{\prime} = \frac{d}{2}$.The new fringe width $\beta^{\prime}$ is given by $\beta^{\prime} = \frac{\lambda D^{\prime}}{d^{\prime}}$.Substituting the new values: $\beta^{\prime} = \frac{\lambda (2D)}{d/2} = 4 	imes \frac{\lambda D}{d}$.Therefore,$\beta^{\prime} = 4\beta$.The fringe width becomes $4$ times the original value.

In Young's double slit experiment,if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled,then the fringe width becomes:

Similar Questions

If the source of light used in a Young's double slit experiment is changed from red to violet:

In a double slit experiment,the distance between slits is increased $10$ times whereas their distance from the screen is halved,then what is the fringe width?

In Young's experiment, the intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources is $36 : 1$. The ratio of the amplitudes of the two individual sources will be

In Young's double-slit experiment, the fringe width is $\beta$. If the entire arrangement is placed in a liquid of refractive index $n$, what does the fringe width become?

Vedclass Test Series

Exam Paper Generator

Online Exam Module