A mica strip and a polysterence strip are fitted on the two slite of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polysterene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe- width ? (b) At what distance from the centre will the first maximum be located ?

Explanation

To solve the given problem step by step, we will address both parts (a) and (b) systematically.

### Given Data:
- Thickness of the strips (t) = 0.50 mm = \(0.50 \times 10^{-3}\) m
- Separation between the slits (d) = 0.12 cm = \(0.12 \times 10^{-2}\) m = \(1.2 \times 10^{-3}\) m
- Refractive index of mica (\(μ_m\)) = 1.58
- Refractive index of polystyrene (\(μ_p\)) = 1.55
- Wavelength of light (\(λ\)) = 590 nm = \(590 \times 10^{-9}\) m
- Distance to the screen (D) = 1 m

### Part (a): Calculate the Fringe Width

1. **Formula for Fringe Width**:
The fringe width (\(β\)) in a double slit experiment is given by the formula:
\[ β = \frac{λD}{d} \]

2. **Substituting the Values**:
\[ β = \frac{(590 \times 10^{-9} \, \text{m}) \times (1 \, \text{m})}{(1.2 \times 10^{-3} \, \text{m})} \]

3. **Calculating**:
\[ β = \frac{590 \times 10^{-9}}{1.2 \times 10^{-3}} = 4.91667 \times 10^{-4} \, \text{m} \approx 4.9 \times 10^{-4} \, \text{m} \]

### Part (b): Calculate the Distance of the First Maximum from the Center

1. **Calculate the Optical Path Change**:
The optical path change (\(Δx\)) due to the strips is given by:
\[ Δx = (μ_m - 1)t - (μ_p - 1)t = (μ_m - μ_p)t \]
Substituting the values:
\[ Δx = (1.58 - 1.55) \times (0.50 \times 10^{-3}) = 0.03 \times 0.50 \times 10^{-3} = 0.015 \times 10^{-3} \, \text{m} \]

2. **Calculate the Number of Fringes Shifted**:
The number of fringes shifted (\(n\)) is given by:
\[ n = \frac{Δx}{λ} \]
Substituting the values:
\[ n = \frac{0.015 \times 10^{-3}}{590 \times 10^{-9}} \approx 25.42 \]
This means there are 25 full fringes and a fractional part of 0.42.

3. **Calculate the Position of the First Maximum**:
The position of the first maximum from the center can be calculated as:
\[ x = 0.42 \times β \]
Substituting the values:
\[ x = 0.42 \times (4.9 \times 10^{-4}) \approx 2.058 \times 10^{-4} \, \text{m} \approx 0.02058 \, \text{cm} \approx 0.021 \, \text{cm} \]

### Summary of Results:
- (a) The fringe width \(β\) is approximately \(4.9 \times 10^{-4}\) m.
- (b) The distance of the first maximum from the center is approximately \(0.021\) cm.