Optics Question 231
Question: In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength $ {\lambda _{1}} $ is found to be coincident with the third maximum at $ {\lambda _{2}} $ . So
Options:
A) $ 3{\lambda _{1}}=0.3{\lambda _{2}} $
B) $ 3{\lambda _{1}}={\lambda _{2}} $
C) $ {\lambda _{1}}=3.5{\lambda _{2}} $
D) $ 0.3{\lambda _{1}}=3{\lambda _{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Position of first minima = position of third maxima i.e., $ \frac{1\times {\lambda _{1}}D}{d}=\frac{( 2\times 3+1 )}{2}\frac{{\lambda _{2}}D}{d}\Rightarrow {\lambda _{1}}=3.5{\lambda _{2}} $
