Question

A point source is emitting light of wavelength 6000 $\stackrel{o}{A}$ is placed at a very small height h above a flat reflecting surface $MN$ as shown in the figure. The intensity of the reflected light is $36$% of the incident intensity. Inference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance $D$ from it. If the intensity at $p$ be maximum, then the minimum distance through which the reflecting surface $MN$ should be displaced so that at $P$ again becomes maximum?

• A. 3 $\times$ 10${}^{-7}$ m
• B. 6 $\times$ 10${}^{-7}$ m
• C. 1.5 $\times$ 10${}^{-7}$ m
• D. 12 $\times$ 10${}^{-7}$ m

Answer 1

The correct option is A 3 $\times$ 10${}^{-7}$ m

Light reaches to point P by two ways.

First, directly from source S and second, after reflection from mirror MN. Therefore path difference,

$\Delta x=2h$,

Now, let the surface MN is displaced by $y$.

Hence the path difference,

$\Delta x^{\prime }=2(h\pm y)$ ,

Extra path difference due to displacement of surface MN,

$\Delta x^{\prime \prime }=2(h\pm y)-2h=\pm 2y$

The intensity of light at P will be maximum again, if

$2y=\lambda =6000$,

or $y=6000/2=3000A^o$,

or $y=3\times {10}^{-7}m$