Question

In Young's double slit experiment, white light is used. The separation between the slits is b. The screen is at a distance d (d>> b) from the slits. Some wavelengths are missing exactly in front of one slit. These wavelengths are

[IIT 1984; AIIMS 1995]

• A. $\lambda =\frac{b^2}{d}$
• B. $\lambda =\frac{2b^2}{d}$
• C. $\lambda =\frac{b^2}{3d}$
• D. $\lambda =\frac{2b^2}{3d}$

Answer 1

Answer: (A), (C)

$\lambda =\frac{b^2}{3d}$

Path difference between the rays reaching infront of slit S₁ is:
$S_{1}P-S_{2}P=(b^2+d^2{)}^{1/2}-d$
For distructive interference at P
$S_{1}P-S_{2}P=\frac{(2n-1)\lambda }{2}i.e.,(b^2+d^2{)}^{1/2}-d=\frac{(2n-1)\lambda }{2}\Rightarrow d{(1+\frac{b^2}{d^2})}^{1/2}-d=\frac{(2n-1)\lambda }{2}\Rightarrow d(1+\frac{b^2}{2d^2}+.......)-d=\frac{(2n-1)\lambda }{2}$
(Binomial Expansion)
$\Rightarrow \frac{b^2}{2d}=\frac{(2n-1)\lambda }{2}\Rightarrow \lambda =\frac{b^2}{(2n-1)d}$
For $n=1,\mathrm{2.......},\lambda =\frac{b^2}{d},\frac{b^2}{3d}$