Question 306: Waves 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 the source and screen is ' $D$ '. The wave length of light $(\lambda)$ used is

Accepted Answer

In Young's double-slit experiment, the path difference for the minima in the interference pattern is given by:
$$ d \sin(	heta) = (m + \frac{1}{2}) \lambda $$
where $ m $ is the order of the minimum (with consecutive minima being $ m = 0, 1, 2, \ldots $) and $ 	heta $ is the angle of the fringe with respect to the central line.
Given that the second minimum is observed exactly in front of one slit, the second minimum corresponds to $ m = 1 $. Here, the minima can be assumed symmetrical, and thus directly in front of one slit at $ m = 1 $.
For small angles where $ \sin(	heta) \approx 	an(	heta) \approx \frac{x_m}{D} $, the path difference can be rewritten considering the distance on the screen. Thus:
$$ x_m = \frac{(m + \frac{1}{2}) \lambda D}{d} $$
Given that the minimum occurs directly in front of one slit ($ x_m = \frac{d}{2} $), we plug it in and solve for the wavelength $\lambda$:
$ \frac{d}{2} = \frac{(1 + \frac{1}{2}) \lambda D}{d} $
Solving for $\lambda$:
$ d^2 = (3/2) \lambda D $
$ \lambda = \frac{2d^2}{3D} $
This means the correct option is:
Option C
$$\frac{d^2}{3D}$$

Answer

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

Answer

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

Answer

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

Answer

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

Waves 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 the source and screen is ' D '. The wave length of light (\lambda) used is

Choose the correct answer: