Question

In a Young's double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1m. A parallel beam of light of wavelength 600nm is incident on the slits at angle α as shown in figure. On the screen, the point O is equidistant from the slits and distance PO is 11.0 mm. Which of the following statement(s) is/are correct ?

• A. For $\alpha =\frac{0.36}{\pi }$ degree, there will be destructive interference at point O.
• B. Fringe spacing depends on α
• C. For $\alpha =\frac{0.36}{\pi }$ degree, there will be destructive interference at point P
• D. For α = 0, there will be constructive interference at point P.

Answer 1

Answer: (C)

$\left(1\right)\Delta x=dsin\alpha$

$=d\alpha$ (as $\alpha$ is very small)$\alpha =\frac{.36}{180}=\left(2\times 10^{-3}\right)$rad

$\frac{\Delta x}{\lambda }=\frac{\left(3\times 10^{-4}\right)\left(2\times 10^{-3}\right)}{6\times 10^{-7}}=1$

so constructive interference

$\left(2\right)\beta =\frac{D\lambda }{d}$

$\left(3\right)\Delta x_p=d\alpha +\frac{dy}{D}$

$=3\times 10^{-4}\left(2\times 10^{-3}+11\times 10^{-3}\right)$

$=39\times 10^{-7}$

$\frac{\Delta x_p}{\lambda }=\frac{39\times 10^{-7}}{6\times 10^{-7}}=6.5$ so destructive

$\left(4\right)\Delta x_p=\frac{dy}{D}=\left(3\times 10^{-4}\right)\times 11\times 10^{-3}$

$=33\times 10^{-7}$

$\frac{\Delta x_p}{\lambda }=\frac{33\times 10^{-7}}{6\times 10^{-7}}=5.5\Rightarrow$ destructive