In the Young's double slit experiment,the dis | vedclass

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

Question

(B) The fringe width is given by $\beta = \frac{\lambda D}{d}$.For the fringe width to be maximum,the slit separation $d$ must be minimum,and for the

Vedclass · Accepted Answer

$\frac{2 \lambda D a_{0}}{d_{0}^{2}-a_{0}^{2}}$

Vedclass · Answer

(B) The fringe width is given by $\beta = \frac{\lambda D}{d}$.For the fringe width to be maximum,the slit separation $d$ must be minimum,and for the fringe width to be minimum,the slit separation $d$ must be maximum.The slit separation is $d(t) = d_{0} + a_{0} \sin \omega t$.The maximum value of $d$ is $d_{\max} = d_{0} + a_{0}$ and the minimum value of $d$ is $d_{\min} = d_{0} - a_{0}$.Therefore,the minimum fringe width is $\beta_{\min} = \frac{\lambda D}{d_{0} + a_{0}}$ and the maximum fringe width is $\beta_{\max} = \frac{\lambda D}{d_{0} - a_{0}}$.The difference between the largest and smallest fringe width is $\beta_{\max} - \beta_{\min} = \frac{\lambda D}{d_{0} - a_{0}} - \frac{\lambda D}{d_{0} + a_{0}}$.Simplifying this expression: $\beta_{\max} - \beta_{\min} = \lambda D \left( \frac{(d_{0} + a_{0}) - (d_{0} - a_{0})}{d_{0}^{2} - a_{0}^{2}} \right) = \frac{2 \lambda D a_{0}}{d_{0}^{2} - a_{0}^{2}}$.

In the Young's double slit experiment,the distance between the slits varies in time as $d(t) = d_{0} + a_{0} \sin \omega t$; where $d_{0}$,$\omega$,and $a_{0}$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as:

Similar Questions

In a double slit experiment,at a certain point on the screen,the path difference between the two interfering waves is $\frac{1}{8}$ of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is:

Monochromatic green light of wavelength $5 \times 10^{-7} \, m$ illuminates a pair of slits $1 \, mm$ apart. The separation of bright lines (fringe width) on the interference pattern formed on a screen $2 \, m$ away is ...... $mm$.

The distance of the $5^{\text{th}}$ dark fringe from the center is $4 \, mm$. If $D = 2 \, m$ and $\lambda = 600 \, nm$,then the distance between the slits is (in $mm$):

What happens to the fringe width in Young's double-slit experiment if it is performed in glycerine instead of air?

In Young's experiment,the distance between slits is $0.28 \,mm$ and the distance between the slits and the screen is $1.4 \,m$. The distance between the central bright fringe and the third bright fringe is $0.9 \,cm$. What is the wavelength of the light used in $\mathring{A}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module