A thin paper of thickness 0.02 mm having a refractive index 1.45 is pasted across one of the slits in a YDSE. The paper transimits `4//9` of the light energy falling on it.
a. Find the ratio of maximum intensity to the minimum intensity in interference pattern.
b. How many fringes will cross through the center if an indentical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.

Explanation

To solve the problem step by step, we will break it down into two parts as given in the question:

### Part (a): Finding the ratio of maximum intensity to minimum intensity in the interference pattern.

1. **Identify the Intensities**:
- Let the intensity of light falling on the first slit be \( I_1 = I \).
- The intensity transmitted through the paper (for the second slit) is given as \( I_2 = \frac{4}{9} I_1 = \frac{4}{9} I \).

2. **Calculate the Ratio of Intensities**:
- The ratio of the intensities is given by:
\[ \frac{I_1}{I_2} = \frac{I}{\frac{4}{9} I} = \frac{9}{4} \]

3. **Relate the Intensities to Amplitudes**:
- Since intensity is proportional to the square of the amplitude, we can write:
\[ \frac{I_1}{I_2} = \left(\frac{A_1}{A_2}\right)^2 \]
- Therefore, we have:
\[ \frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{9}{4}} = \frac{3}{2} \]

4. **Calculate Maximum and Minimum Intensities**:
- The maximum intensity \( I_{max} \) and minimum intensity \( I_{min} \) in the interference pattern can be expressed as:
\[ I_{max} = (A_1 + A_2)^2 \quad \text{and} \quad I_{min} = (A_1 - A_2)^2 \]
- Substituting \( A_1 = 3k \) and \( A_2 = 2k \):
\[ I_{max} = (3k + 2k)^2 = (5k)^2 = 25k^2 \]
\[ I_{min} = (3k - 2k)^2 = (k)^2 = k^2 \]

5. **Find the Ratio of Maximum to Minimum Intensity**:
- The ratio is:
\[ \frac{I_{max}}{I_{min}} = \frac{25k^2}{k^2} = 25 \]
- Thus, the ratio of maximum intensity to minimum intensity is:
\[ \frac{I_{max}}{I_{min}} = 25:1 \]

### Part (b): Finding the number of fringes that will cross through the center.

1. **Identify the Variables**:
- The refractive index \( \mu = 1.45 \)
- Thickness of the paper \( T = 0.02 \, \text{mm} = 0.02 \times 10^{-3} \, \text{m} \)
- Wavelength \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)

2. **Calculate the Optical Path Difference**:
- The optical path difference introduced by the paper is given by:
\[ \Delta = (\mu - 1) T \]
- Substituting the values:
\[ \Delta = (1.45 - 1) \times 0.02 \times 10^{-3} = 0.45 \times 0.02 \times 10^{-3} = 9 \times 10^{-6} \, \text{m} \]

3. **Calculate the Number of Fringes**:
- The number of fringes \( n \) that will cross through the center is given by:
\[ n = \frac{\Delta}{\lambda} \]
- Substituting the values:
\[ n = \frac{9 \times 10^{-6}}{600 \times 10^{-9}} = \frac{9}{0.6} = 15 \]

### Final Answers:
- (a) The ratio of maximum intensity to minimum intensity is \( 25:1 \).
- (b) The number of fringes that will cross through the center is \( 15 \).