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
Show Answer
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 $ .
