Question 232: Wave Optics

Question

In Young's double slit experiment, in an interference pattern second minimum is observed exactly in front of one slit. The distance between the two coherent sources is ' $d$ ' and the distance between source and screen is ' $D$ '. The wavelength of light source used is

Accepted Answer

Step 1: Path difference at a point on screen
Suppose the slits are $S_1$ and $S_2$, separated by distance $d$. Consider the point $P$ on the screen which is exactly in front of slit $S_2$. Then:
Path from $S_2$ to $P$ = $D$.
Path from $S_1$ to $P$: slightly longer because it is displaced by $d$.
Using geometry, the path difference is approximately
$ \Delta = \frac{d \cdot x}{D} $
where $x$ is the distance of point $P$ from the central axis (the midpoint between slits projected onto screen).
But here point $P$ is directly in front of a slit (say slit $S_2$). The distance of that point on screen from the central axis is simply $x = 	frac{d}{2}$.
(Reason: midpoint between slits is at $x=0$; each slit is $\pm 	frac{d}{2}$ from that axis; projecting forward keeps that offset).
Thus,
$ \Delta = \frac{d \cdot (d/2)}{D} = \frac{d^2}{2D}. $
Step 2: Condition for minimum
Nth
minimum
condition in YoungÃ¢ÂÂs setup:
$ \Delta = \Big(m + 	frac{1}{2}\Big)\lambda, \quad m=0,1,2,... $
The
second minimum
means $m=1$ (since $m=0$ is first minimum).
So:
$ \Delta = \frac{3}{2}\lambda. $
Step 3: Equating path difference
$ \frac{d^2}{2D} = \frac{3}{2}\lambda $
$ \lambda = \frac{d^2}{3D}. $
Ã¢ÂÂ Final Answer:
$ \boxed{	frac{d^2}{3D}} $
That corresponds to
Option B
.

Answer

$\frac{d^2}{4 D}$

Answer

$\frac{d^2}{3 D}$

Answer

$\frac{d^2}{2 D}$

Answer

$\frac{d^2}{D}$

Wave Optics

In Young's double slit experiment, in an interference pattern second minimum is observed exactly in front of one slit. The distance between the two coherent sources is ' d ' and the distance between source and screen is ' D '. The wavelength of light source used is

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