Question

A point source $S$ emits light of wavelength $600\text{nm}.$ It is placed at a very small distance from a flat reflecting mirror $\text{AB}$. Interference fringes are observed on the screen placed parallel to the reflecting surface at very large distance $D$ from it. The ratio of minimum to maximum intensities in the interference fringes formed near the point $\text{P}$ is $\frac{1}{6}.$ If the ratio of intensity of the reflected light to the intensity of incident light is $0.09k.$ The value of $k$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

According to the problem


$\frac{I_{min}}{I_{max}}=\frac{1}{6}$

$\Rightarrow \frac{(\sqrt{I}-{\sqrt{I}}^{\prime }{)}^2}{(\sqrt{I}+{\sqrt{I}}^{\prime }{)}^2}=\frac{1}{6}$

$\Rightarrow \frac{\sqrt{I}-{\sqrt{I}}^{\prime }}{\sqrt{I}+\sqrt{I^{\prime }}}=\frac{1}{\sqrt{6}}$

Let, $\sqrt{\frac{I}{I^{\prime }}}=a$
So,
$\frac{\sqrt{I}-\sqrt{I^{\prime }}}{\sqrt{I}+\sqrt{I^{\prime }}}=\frac{1}{\sqrt{6}}=\frac{a-1}{a+1}$

$\Rightarrow \sqrt{6}a-\sqrt{6}=a+1$

$\Rightarrow (\sqrt{6}-1)a=(\sqrt{6}+1)$

$\Rightarrow \frac{\sqrt{6}-1}{\sqrt{6}+1}=\frac{1}{a}=\sqrt{\frac{I^{\prime }}{I}}$

$\Rightarrow \frac{I^{\prime }}{I}={\left(\frac{\sqrt{6}-1}{\sqrt{6}+1}\right)}^2={\left(\frac{2.49-1}{2.49+1}\right)}^2$

$\Rightarrow \frac{I^{\prime }}{I}=0.18$

Comparing with $\frac{I^{\prime }}{I}=0.09k$

$\Rightarrow k=2$

Hence, option $\left(\text{B}\right)$ is correct.