Optics Question 867
Question: In young’s double - slit experiment, the separation between the slits is d, distance between the slit and screen is D (D»d). In the interference pattern, there is a maxima exactly in front of each slit. Then the possible wavelength(s) used in the experiment are
Options:
A) $ d^{2}/D,d^{2}/2D,d^{2}/3D $
B) $ d^{2}/D,d^{2}/3D,d^{2}/5D $
C) $ d^{2}/2D,d^{2}/4D,d^{2}/6D $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ S _{2}P-S _{1}P=\frac{dy}{D}=\frac{d\times (d/2)}{D}=\frac{d^{2}}{2D} $
$ \frac{d^{2}}{2D}=n\lambda $
$ \lambda =\frac{d}{2nD} $ , n=1, 2, ??.. $ \lambda =\frac{d^{2}}{2D},\frac{d^{2}}{4D},\frac{d^{2}}{6D} $
