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.Calculate the ratio of the minimum to the maximum intensities in the interference fringes formed near the point $P$ (shown in the figure).

Answer 1

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

$\frac{I_{max}}{I_{min}}=\frac{(\sqrt{I_1}+\sqrt{I_2}{)}^2}{({\sqrt{I}}_1-\sqrt{I_2}{)}^2}=\frac{(\sqrt{I}+\sqrt{0}.36)I^2}{(\sqrt{I}-\sqrt{0}.36)I^2}=\frac{1}{16}=0.0625$

$\text{Hence 0.0625 is the correct answer}$