Question

A point sources $S$ emitting light of wavelength $600nm$ is placed at a very small height $h$ above a flat reflecting surface $AB$ (see figure). The intensity of the reflected light is $36\%$ of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance $D$ from it. If the intensity at [pint $P$ corresponds to a maximum, calculate the minimum distance(in nm) through which the reflecting surface $AB$ should be shifted so that the intensity at $P$ again becomes maximum.

Answer 1

$\text{Given}:$ $\lambda =600nm$, I(reflected) = 36 % I(incident)

$\text{Solution}$

The path difference between the rays 1 and 2 coming from S and S' will be equal on the circular path on the screen, hence fringes will be circular.

Let the intensity of light coming from S be $I_1=I$, then as per the problem, the velocity of the reflected light will be $I_2=0.36I$.

As we know

$I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2$

$I_{min}=(\sqrt{I_1}-\sqrt{I_2}{)}^2$

Hence,

$\frac{I_{max}}{I_{min}}=\frac{(\sqrt{I_1}+{\sqrt{I_2}}^2)}{({\sqrt{I}}_1-\sqrt{I_2}{)}^2}=\frac{(\sqrt{I}+\sqrt{0}.36I{)}^2}{(\sqrt{I}-\sqrt{0}.36I{)}^2}=\frac{1}{16}$

To have the next maximum at P, the path difference between the interfering waves must change by $\lambda$.

If AB is moved by a distance x, it will cause an additional path difference 2x

$2x=\lambda$ (value minimum value of x)

$⟹x=\frac{\lambda }{2}=300mm$

$\text{Hence the correct answeer is 300 mm}$