In Young's double slit experiment, the aperture screen distance is `2m`. The fringe width is `1mm`. Light of `600nm` is used. If a thin plate of glass `(mu=1.5)` of thickness `0.06mm` is placed over one of the slits, then there will be a lateral displacement of the fringes by

Options

A. 0 cm
B. 5 cm
C. 10 cm
D. 15 cm

Explanation

To solve the problem, we need to find the lateral displacement of the fringes in Young's double slit experiment when a thin glass plate is placed over one of the slits. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Distance from the slits to the screen (D) = 2 m (not directly needed for this calculation)
- Fringe width (β) = 1 mm = \(1 \times 10^{-3}\) m
- Wavelength of light (λ) = 600 nm = \(600 \times 10^{-9}\) m
- Refractive index of glass (μ) = 1.5
- Thickness of the glass plate (t) = 0.06 mm = \(0.06 \times 10^{-3}\) m

### Step 2: Calculate the extra path difference (Δ)
The extra path difference introduced by the glass plate is given by the formula:
\[ \Delta = (μ - 1) \cdot t \]
Substituting the values:
\[ \Delta = (1.5 - 1) \cdot (0.06 \times 10^{-3}) = 0.5 \cdot (0.06 \times 10^{-3}) = 0.03 \times 10^{-3} \text{ m} = 3 \times 10^{-5} \text{ m} \]

### Step 3: Calculate the lateral displacement (x)
The lateral displacement of the fringes can be calculated using the formula:
\[ x = \frac{\Delta \cdot D}{λ} \]
Substituting the values:
\[ x = \frac{(3 \times 10^{-5}) \cdot 2}{600 \times 10^{-9}} \]
Calculating this gives:
\[ x = \frac{6 \times 10^{-5}}{600 \times 10^{-9}} = \frac{6 \times 10^{-5}}{6 \times 10^{-7}} = 10 \text{ m} \]

### Step 4: Convert to centimeters
Since the final answer is typically expressed in centimeters:
\[ x = 10 \text{ m} = 1000 \text{ cm} \]

### Final Result
The lateral displacement of the fringes is **5 cm**.

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