In a double slit experiment the distance between the slits is $0.1\ ext{cm}$ and the screen is placed at $50\ ext{cm}$ from the slits plane. When one slit is covered with a transparent sheet having thickness $𝑡$ and refractive index $𝑛(=1.5)$, the central fringe shifts by $0.2\ ext{cm}$. The value of $𝑡$ is
_____ cm.

Question

In a double slit experiment the distance between the slits is $0.1\ 	ext{cm}$ and the screen is placed at $50\ 	ext{cm}$ from the slits plane. When one slit is covered with a transparent sheet having thickness $𝑡$ and refractive index $𝑛(=1.5)$, the central fringe shifts by $0.2\ 	ext{cm}$. The value of $𝑡$ is
_____ cm.

Accepted Answer

B. 1. When a transparent sheet of thickness $t$ and refractive index $n$ is placed in front of one slit, the light travels more slowly inside the sheet than it would in air.

The additional optical path difference introduced is

$$
(n-1)t.
$$

2. In Young's double slit experiment, the geometric path difference for a point at a distance $y$ from the central axis is

$$
\frac{dy}{D},
$$

where $d$ is the slit separation and $D$ is the distance of the screen from the slits.

3. The shifted central fringe is the position where the additional optical path difference is exactly balanced by the geometric path difference.

Therefore,

$$
\begin{aligned}
\frac{dy}{D}
&=(n-1)t.
\end{aligned}
$$

4. Rearranging,

$$
\begin{aligned}
t
&=\frac{yd}{D(n-1)}.
\end{aligned}
$$

5. Substituting the given values,

$$
y=0.2\ 	ext{cm},\quad
d=0.1\ 	ext{cm},\quad
D=50\ 	ext{cm},\quad
n=1.5,
$$

$$
\begin{aligned}
t
&=\frac{0.2	imes0.1}{50(1.5-1)} \
&=\frac{0.02}{25} \
&=8	imes10^{-4}\ 	ext{cm}.
\end{aligned}
$$

6. Hence, the required thickness of the sheet is

$$
\boxed{8	imes10^{-4}\ 	ext{cm}},
$$

so the correct option is **B**.