A transparent paper sheet of thickness 0.03 mm and refractive index 1.45 is placed in front of one of the slits in a double slit experiment. The paper transmits 9/25 of the light energy falling on it. If the wavelength of light used is 4500 Å . What is the ratio of maximum to minimum intensity in the interference pattern.

Explanation

To solve the problem, we need to determine the ratio of maximum to minimum intensity in the interference pattern created by a double-slit experiment, where one of the slits has a transparent paper in front of it. Here's a step-by-step solution:

### Step 1: Convert Given Values
- **Thickness of the paper (t)**: 0.03 mm = \(0.03 \times 10^{-3}\) m = \(3 \times 10^{-5}\) m
- **Refractive index (n)**: 1.45
- **Wavelength of light (\(\lambda\))**: 4500 Å = \(4500 \times 10^{-10}\) m = \(4.5 \times 10^{-7}\) m

### Step 2: Calculate the Optical Path Difference
...