The figure shows a Young's double slit experimental setup. It is observed that when a thin transparent sheet of thickness t and refractive index $$\mu$$ is put in front of one of the slits, the central maximum gets shifted by a distance equal to n fringe width. If the wavelength of light used is $$\lambda$$ then t will be:

Step-by-Step Derivation

1. Shift due to a Transparent Sheet

When a transparent sheet of thickness $$t$$ and refractive index $$\mu$$ is introduced in front of one of the slits, it introduces an additional optical path difference ($$\Delta x$$) given by:

$$\Delta x = (\mu - 1)t$$

Due to this path difference, the entire fringe pattern shifts on the screen. The linear shift of the central maximum ($$\Delta y$$) is given by the formula:

$$\Delta y = \frac{D}{a}(\mu - 1)t$$

Where:

• $$D$$ = distance between the slits and the screen
• $$a$$ = distance between the two slits

2. Relating the Shift to Fringe Width

The width of a single fringe ($$\beta$$) in Young's double slit experiment is defined as:

$$\beta = \frac{D\lambda}{a}$$

According to the problem statement, the central maximum shifts by a distance equal to $$n$$ fringe widths:

$$\Delta y = n\beta$$

3. Equating and Finding Thickness ($$t$$)

Substitute the expressions for $$\Delta y$$ and $$\beta$$ into the equation:

$$\frac{D}{a}(\mu - 1)t = n \left(\frac{D\lambda}{a}\right)$$

We can cancel out the common terms $$\frac{D}{a}$$ from both sides of the equation:

$$(\mu - 1)t = n\lambda$$

Solving for the thickness $$t$$:

$$t = \frac{n\lambda}{\mu - 1}$$

Correct Answer

The correct option is (2) $$\frac{n\lambda}{\mu - 1}$$.