A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed ?

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the Given Data
We are given:
- Refractive index (μ) = 1.45
- Thickness of the paper (t) = 0.02 mm
- Wavelength of light (λ) = 620 nm

### Step 2: Convert Units
Convert the thickness from millimeters to meters:
- \( t = 0.02 \, \text{mm} = 0.02 \times 10^{-3} \, \text{m} \)

Convert the wavelength from nanometers to meters:
- \( \lambda = 620 \, \text{nm} = 620 \times 10^{-9} \, \text{m} \)

### Step 3: Calculate the Change in Optical Path
The change in optical path due to the transparent paper is given by:
\[ \Delta = (μ - 1) \cdot t \]
Substituting the values:
\[ \Delta = (1.45 - 1) \cdot (0.02 \times 10^{-3}) \]
\[ \Delta = 0.45 \cdot (0.02 \times 10^{-3}) \]
\[ \Delta = 0.009 \times 10^{-3} \, \text{m} = 9 \times 10^{-6} \, \text{m} \]

### Step 4: Determine the Number of Fringes
The number of fringes (n) that will cross through the center when the paper is removed is given by:
\[ n = \frac{\Delta}{\lambda} \]
Substituting the values:
\[ n = \frac{9 \times 10^{-6}}{620 \times 10^{-9}} \]

### Step 5: Calculate n
Now, perform the calculation:
\[ n = \frac{9 \times 10^{-6}}{620 \times 10^{-9}} = \frac{9}{620} \times 10^{3} \]
\[ n \approx 14.52 \]

Since the number of fringes must be a whole number, we round it to the nearest whole number:
\[ n \approx 14.5 \]

### Final Answer
The number of fringes that will cross through the center when the paper is removed is approximately 14.5.

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