A young's double slit apporatus is immersed in a liquid of refractive index 1.33.It has slit separation of 1 mm and interference pattern is observed on the screen at a distance 1.33 m from plane of slits.The wavelength in air is `6300 Å`
One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to interchange the position of minima and maxima .

Options

A. 2.57 mm
B. 1.57 mm
C. 3.27 mm
D. 4.18 mm

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript while ensuring clarity and completeness.

### Step 1: Understand the Problem
We have a Young's double slit apparatus immersed in a liquid with a refractive index of 1.33. The slit separation is 1 mm, and the wavelength of light in air is 6300 Å (or 6300 x 10^-10 m). One of the slits is covered with a thin glass sheet of refractive index 1.53. We need to find the smallest thickness of the glass sheet that will interchange the positions of minima and maxima in the interference pattern.

### Step 2: Determine the Effective Wavelength in the Liquid
The wavelength of light in a medium is given by:
\[ \lambda' = \frac{\lambda}{\mu} \]
where \(\lambda\) is the wavelength in air and \(\mu\) is the refractive index of the medium.

Given:
- \(\lambda = 6300 \, \text{Å} = 6300 \times 10^{-10} \, \text{m}\)
- \(\mu = 1.33\)

Calculating the effective wavelength in the liquid:
\[ \lambda' = \frac{6300 \times 10^{-10}}{1.33} \approx 4.73 \times 10^{-10} \, \text{m} \]

### Step 3: Calculate the Relative Refractive Index
The relative refractive index of the glass sheet with respect to the liquid is calculated as:
\[ \mu_{relative} = \frac{\mu_{glass}}{\mu_{liquid}} - 1 \]
where:
- \(\mu_{glass} = 1.53\)
- \(\mu_{liquid} = 1.33\)

Calculating:
\[ \mu_{relative} = \frac{1.53}{1.33} - 1 \approx 0.1511 \]

### Step 4: Set Up the Equation for Thickness
To interchange the positions of minima and maxima, the optical path difference introduced by the glass sheet must be equal to \(\frac{\lambda'}{2}\):
\[ \mu_{relative} \cdot t = \frac{\lambda'}{2} \]
Substituting for \(\lambda'\):
\[ \left(\frac{1.53}{1.33} - 1\right) t = \frac{4.73 \times 10^{-10}}{2} \]

### Step 5: Solve for Thickness \(t\)
Rearranging the equation to find \(t\):
\[ t = \frac{\frac{4.73 \times 10^{-10}}{2}}{\mu_{relative}} = \frac{2.365 \times 10^{-10}}{0.1511} \]

Calculating \(t\):
\[ t \approx 1.57 \times 10^{-10} \, \text{m} = 1.575 \, \mu m \]

### Final Answer
The smallest thickness of the glass sheet required to interchange the positions of minima and maxima is approximately **1.575 micrometers**.

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