Question

In the diagram shown, the separation between the slit is equal to $3\lambda$, where $\lambda$ is the wavelength of the light incident on the plane of the slits. A thin film of thickness $3\lambda$ & refractive index $2$ has been placed in the front of the upper slit. The distance of the central maxima on the screen from $O$ is-

• A. $\frac{\lambda d}{D}$
• B. $\frac{\lambda D}{d}$
• C. $1.5D$
• D. $D$

Answer 1

The correct option is D $D$


When a thin film of thickness $t\&$ refractive index $\mu$ is placed in front of a source, then there is a shift in the position of the fringes.

The position of $n^{th}$ maxima is given by,

$\begin{array}{c}{y}_{nth}=\frac{n\lambda D}{d}-\underset{\text{fringe shift}}{\underset{}{(\mu -1)t\frac{D}{d}}}\end{array}$

Given, $d=3\lambda ;t=3\lambda ;\mu =2$

Now, the Shift in the position of the central maxima $\left(\Delta y\right)$ is given by,

$\Delta y=(\mu -1)t\frac{D}{d}=\frac{(2-1)\times 3\lambda \times D}{3\lambda }$

$\therefore \Delta y=D$

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

Why this Question ? Key concept: When a thin film of thickness $t\&$ refractive index $\mu$ is placed in front of any one of a source, The fringe pattern gets displaced towards the beam in whose path the sheet is introduced. This shift is known as lateral displacement or lateral shift. Lateral shift is given by, $y=(\mu -1)t\frac{D}{d}$