Optics Question 140
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]
Options:
A) $ \lambda =\frac{b^{2}}{d} $
B) $ \lambda =\frac{2b^{2}}{d} $
C) $ \lambda =\frac{b^{2}}{3d} $
D) $ \lambda =\frac{2b^{2}}{3d} $
Show Answer
Answer:
Correct Answer: A
Solution:
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}{2d}=\frac{(2n-1)\lambda }{2}\Rightarrow \lambda =\frac{b^{2}}{(2n-1)d} $ For $ n=1,\ 2…………,\ \lambda =\frac{b^{2}}{d},\ \frac{b^{2}}{3d} $
