Question 267: Wave Optics

Question

In Young's double slit experiment let 'd' be the distance between two slits and 'D' be the distance between the slits and the screen. Using a monochromatic source of wavelength ' $\lambda$ ', in an interference pattern, third minimum is observed exactly in front of one of the slits. If at the same point on the screen first minimum is to be obtained, the required change in the wavelength is [ $\mathrm{d} \& \mathrm{D}$ are not changed].

Accepted Answer

Let the
distance between slits
be $d$,
Distance from slits to screen
be $D$,
Wavelength
is $\lambda$.
Let
point P
on screen be directly in front of the
second slit, S_2
.
1. Path difference at point P
At point P (just in front of $S_2$):
Distance from $S_1$ to P = $d$ farther compared to $S_2$ to P
So,
Path difference
$= d$.
2. Condition for $n^{	ext{th}}$ Minimum in YDSE
The position of the
$n^{	ext{th}}$ minimum
from the central bright is given by:
$ 	ext{Path difference for $n^{th}$ minimum} = (2n-1)\frac{\lambda}{2} $
3. For third minimum ($n=3$) at P :
Set path difference $= d$:
$ d = (2 	imes 3 - 1) \frac{\lambda}{2} $
$ d = (6 - 1) \frac{\lambda}{2} $
$ d = \frac{5\lambda}{2} $
4. For first minimum at the
same point P
with new wavelength ($\lambda'$):
$ d = (2 	imes 1 - 1) \frac{\lambda'}{2} $
$ d = (2-1)\frac{\lambda'}{2} $
$ d = \frac{\lambda'}{2} $
5. Equate both values of $d$ (slit separation not changed):
$ \frac{5\lambda}{2} = \frac{\lambda'}{2} $
$ 5\lambda = \lambda' $
6. Required Change in Wavelength
Change in wavelength:
$ \Delta\lambda = \lambda' - \lambda = 5\lambda - \lambda = 4\lambda $
Final Answer
$\boxed{4\lambda}$
(Correct option:
C
)

Answer

$2 \lambda$

Answer

$3 \lambda$

Answer

$4 \lambda$

Answer

$5 \lambda$

Wave Optics

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