Optics Question 169
Question: In a single slit diffraction experiment first minimum for red light (660 nm) coincides with first maximum of some other wavelength l’. The value of l’ is
Options:
A) $ 4400{AA} $
B) $ 6600{AA} $
C) $ 2000{AA} $
D) $ 3500{AA} $
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Answer:
Correct Answer: A
Solution:
In a single slit diffraction experiment, position of minima is given by $ d\sin \theta =n\lambda $
So for first minima of red $ \sin \theta =1\times ( \frac{{\lambda _{R}}}{d} ) $
and as first maxima is midway between first and second minima, for wavelength $ {\lambda }’ $ , its position will be $ d\sin {\theta }’=\frac{{\lambda }’+2{\lambda }’}{2}\Rightarrow \sin {\theta }’=\frac{3{\lambda }’}{2d} $
According to given condition $ \sin \theta =\sin {\theta }’ $
$ \Rightarrow {\lambda }’=\frac{2}{3}{\lambda _{R}} $ so $ {\lambda }’=\frac{2}{3}\times 6600=440nm $
$ =4400{AA} $
BETA
