A plane monochromatic light wave falls normally on a diaphragm with two narrow slits separated by 2.5mm. The fringe pattern is formed on a screen 100 cm behind the diaphragm. By what distance will these fringes be displaced, when one of the slits is covered by a glass plate ` (mu = 1.5)` of thickness `10 mu m ?`

Options

A. 2mm
B. 1mm
C. 3mm
D. 4mm

Explanation

To solve the problem step by step, we will calculate the displacement of the fringe pattern when one of the slits is covered by a glass plate.

### Step 1: Calculate the optical path difference introduced by the glass plate.

The formula for the optical path difference (Δ) when a glass plate of thickness \( T \) and refractive index \( \mu \) is introduced is given by:

\[ \Delta = (μ - 1) \cdot T \]

Given:
- \( μ = 1.5 \)
- \( T = 10 \, \mu m = 10 \times 10^{-6} \, m \)

Substituting the values:

\[ \Delta = (1.5 - 1) \cdot (10 \times 10^{-6}) = 0.5 \cdot (10 \times 10^{-6}) = 5 \times 10^{-6} \, m \]

### Step 2: Calculate the displacement of the fringe pattern on the screen.

The displacement of the fringe pattern (Δx) on the screen can be calculated using the formula:

\[ \Delta x = \frac{\Delta \cdot D}{d} \]

Where:
- \( D \) = distance from the slits to the screen = 100 cm = 1 m
- \( d \) = separation between the slits = 2.5 mm = 2.5 \times 10^{-3} \, m

Substituting the values:

\[ \Delta x = \frac{5 \times 10^{-6} \cdot 1}{2.5 \times 10^{-3}} = \frac{5 \times 10^{-6}}{2.5 \times 10^{-3}} = 2 \times 10^{-3} \, m \]

### Step 3: Convert the displacement into millimeters.

Since \( 1 \, m = 1000 \, mm \):

\[ \Delta x = 2 \times 10^{-3} \, m = 2 \, mm \]

### Final Answer:
The fringes will be displaced by **2 mm**.

---