Optics Question 851

Question: Monochromatic light of wavelength 400 nm and 560 nm are incident simultaneously and normally on double slits apparatus whose slits separation is 0.1 mm and screen distance is 1m. Distance between areas of total darkness will be

Options:

A) 4mm

B) 5.6 mm

C) 14mm

D) 28mm

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Answer:

Correct Answer: D

Solution:

[d] At the area of total darkness minima will occur for both the wavelengths.

$ \therefore \frac{(2n+1)}{2}{\lambda _{1}}=\frac{(2m+1)}{2}{\lambda _{2}} $

$ \Rightarrow (2n+1){\lambda _{1}}=(2m+1){\lambda _{2}} $

or $ \frac{(2n+1)}{(2m+1)}=\frac{560}{400}=\frac{7}{5} $ or $ 10n=14m+2 $ by inspection

for m=2, n=3 and for m=7, n= 10, the distance between them will be the distance between such points.

i.e., $ \Delta s=\frac{D{\lambda _{1}}}{d}{ \frac{(2n _{2}+1)-(2n _{1}+1)}{2} } $

put $ n _{2}=10 $ , $ n _{1}=3 $ On solving we get, $ \Delta s=28mm $ .



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