Optics Question 151
Question: In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by $ 5\times {{10}^{-2}}m $ towards the slits, the change in fringe width is $ 3\times {{10}^{-5}}m $ . If separation between the slits is $ {{10}^{-3}}m $ , the wavelength of light used is
[Roorkee 1992]
Options:
A) $ 6000{AA} $
B) $ 5000{AA} $
C) $ 3000{AA} $
D) $ 4500{AA} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \beta =\frac{\lambda D}{d} $
Therefore b µ D
Therefore $ \frac{{\beta _{1}}}{{\beta _{2}}}=\frac{D _{1}}{D _{2}} $
Therefore $ \frac{{\beta _{1}}-{\beta _{2}}}{{\beta _{2}}}=\frac{D _{1}-D _{2}}{D _{2}} $
Therefore $ \frac{\Delta \beta }{\Delta D}=\frac{{\beta _{2}}}{D _{2}}=\frac{{\lambda _{2}}}{d _{2}} $
$ ={\lambda _{2}}=\frac{3\times {{10}^{-5}}}{5\times {{10}^{-2}}}\times {{10}^{-3}}=6\times {{10}^{-7}}m=6000{AA} $
BETA
