Optics Question 156
Question: In a Young’s double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength $ {\lambda _{0}}=750nm $ and $ \lambda =900nm $ . The minimum distance from the common central bright fringe on a screen 2m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is
Options:
A) 1.5 mm
B) 3 mm
C) 4.5 mm
D) 6 mm
Show Answer
Answer:
Correct Answer: C
Solution:
From the given data, note that the fringe width (b1) for $ {\lambda _{1}}=900nm $ is greater than fringe width (b2) for $ {\lambda _{2}}=750nm. $
This means that at though the central maxima of the two coincide, but first maximum for $ {\lambda _{1}}=900nm $ will be further away from the first maxima for $ {\lambda _{2}}=750nm, $ and so on.
A stage may come when this mismatch equals b2, then again maxima of $ {\lambda _{1}}=900nm, $ will coincide with a maxima of $ {\lambda _{2}}=750nm, $
let this correspond to nth order fringe for l1. Then it will correspond to $ {{(n+1)}^{th}} $ order fringe for l2.
Therefore $ \frac{n{\lambda _{1}}D}{d}=\frac{(n+1){\lambda _{2}}D}{d} $
$ \Rightarrow n\times 900\times {{10}^{-9}}=(n+1)750\times {{10}^{-9}}\Rightarrow n=5 $
Minimum distance from Central maxima $ =\frac{n{\lambda _{1}}D}{d}=\frac{5\times 900\times {{10}^{-9}}\times 2}{2\times {{10}^{-3}}} $
$ =45\times {{10}^{-4}}m=4.5mm $
