Consider a Young's double slit experiment as | vedclass

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

Question

(A) For the first minima to occur at point $P$ (directly in front of $S_1$),the path difference between the waves from $S_1$ and $S_2$ must be equal t

Vedclass · Accepted Answer

$\frac{\lambda}{2(\sqrt{5}-2)}$

Vedclass · Answer

(A) For the first minima to occur at point $P$ (directly in front of $S_1$),the path difference between the waves from $S_1$ and $S_2$ must be equal to $\frac{\lambda}{2}$.The distance from $S_1$ to $P$ is $2d$.The distance from $S_2$ to $P$ is $\sqrt{(2d)^2 + d^2} = \sqrt{4d^2 + d^2} = \sqrt{5}d$.The path difference $\Delta x = S_2P - S_1P = \sqrt{5}d - 2d = d(\sqrt{5} - 2)$.Setting the path difference equal to $\frac{\lambda}{2}$ for the first minima:$d(\sqrt{5} - 2) = \frac{\lambda}{2}$Therefore,the slit separation $d$ is:$d = \frac{\lambda}{2(\sqrt{5} - 2)}$

Consider a Young's double slit experiment as shown in the figure. What should be the slit separation $d$ in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit $S_1$?

Similar Questions

In a $YDSE$,if the slits are of unequal width:

In a Young's double slit experiment,the slit separation is $0.2 \ cm$,the distance between the screen and slit is $1 \ m$. The wavelength of the light used is $5000 \ \mathring{A}$. The distance between two consecutive dark fringes (in $mm$) is

The fringe width for red colour as compared to that for violet colour is approximately (in $times$)

In a Young's double-slit experiment,the wavelength of light used is $6000 \, \mathring A$. The phase difference between the central bright fringe and the third bright fringe is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module