Optics Question 1091
Question: In Young’s double slit experiment, the fringe width is $ 1\times {{10}^{-4}}m $ if the distance between the slit and screen is doubled and the distance between the two slit is reduced to half and wavelength is changed from $ 6.4\times 10^{7}m $ to $ 4.0\times {{10}^{-7}}m $ , the value of new fringe width will be
Options:
A) $ 0.15\times {{10}^{-4}}m $
B) $ 2.0\times {{10}^{-4}}m $
C) $ 1.25\times {{10}^{-4}}m $
D) $ 2.5\times {{10}^{-4}}m $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \beta =\frac{\lambda D}{d} $
$ \Rightarrow \frac{{\beta _{2}}}{{\beta _{1}}}=\frac{{\lambda _{2}}D _{2}d _{1}}{{\lambda _{1}}D _{1}d _{2}} $
$ \Rightarrow {\beta _{2}}=2.5\times {{10}^{-4}}m $ .
BETA
