Question

${\text{M}}_1$ and ${\text{M}}_2$ are two plane mirrors which are kept parallel to each other as shown. There is a point $\text{O}$ on the perpendicular screen just in front of $\text{S}.$ What should be the wavelength of light coming from monochromatic source $\text{S},$ so that a maxima is formed at $\text{O}$ due to interference of reflected light from both the mirrors. [consider only 1st reflection and $D>>d,d>>\lambda$]

• A. $\frac{3d^2}{D}$
• B. $\frac{3d^2}{2D}$
• C. $\frac{d^2}{D}$
• D. $\frac{2d^2}{D}$

Answer 1

The correct option is B $\frac{3d^2}{2D}$


As the property of plane mirror, distance of image is equal to distance of objective. Thus, image of $\text{S}$ due to ${\text{M}}_1$ and ${\text{M}}_2$ acts as a two coherent sources ${\text{S}}_1$ and ${\text{S}}_2$ respectively.



From the figure, let point $\text{P}$ be the central maxima and point $\text{O}$ be $n^{th}$ maxima.

Separation between $\text{O}$ and $\text{P}$ is $y=\frac{d}{2}$ and separation between ${\text{S}}_1$ and ${\text{S}}_2$ is $d^{\prime }=3d$

Now, for the point $\text{O}$ be a maxima,

$\Delta x=\frac{y{d}^{\prime }}{D}=n\lambda$

$n\lambda =\frac{\frac{d}{2}\times 3d}{D}=\frac{3d^2}{2D}$

$\lambda =\frac{3d^2}{2Dn}$

$\lambda =\frac{3d^2}{2D}[\text{If}n=1]$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.