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)$\quad\quad\quad\alpha = \frac{.36}{180} = \left(2\times10^{-3}\right)$rad

$\frac{\Delta x}{\lambda} = \frac{\left(3\times10^{-4}\right) \left(2\times10^{-3}\right)}{6\times10^{-7}} = 1$

so constructive interference

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

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

$= 3 \times 10^{-4} \left(2\times10^{-3}+11 \times10^{-3}\right)$

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

$\frac{\Delta x_{p}}{\lambda} = \frac{39\times10^{-7}}{6\times10^{-7}} = 6.5$ so destructive

$\left(4\right)\,\Delta x_{p} = \frac{dy}{D} = \left(3\times10^{-4}\right)\times11\times10^{-3}$

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

$\frac{\Delta x_{p}}{\lambda} = \frac{33\times10^{-7}}{6\times10^{-7}} = 5.5\,\Rightarrow $ destructive