Interference fringes are produced by a double slit arrangement
and a piece of plane parallel glass of refractive index 1.5 is interposed in one of
the interfering beam. If the fringes are displaced through 30 fringe widths for
light of wavelength `6xx 10^(-5) cm`, find the thickness of the plate.

Explanation

To solve the problem, we need to find the thickness of the glass plate that causes a displacement of the interference fringes. Here’s a step-by-step solution:

### Step 1: Understand the relationship between fringe displacement and path difference
The displacement of the interference fringes due to the introduction of a glass plate is related to the path difference introduced by the plate. The path difference (Δ) caused by the glass slab can be expressed as:
\[ \Delta = (n - 1) \cdot t \]
where \( n \) is the refractive index of the glass and \( t \) is the thickness of the glass.

### Step 2: Relate the path difference to fringe width
The number of fringes displaced (N) is given as 30. The path difference can also be related to the fringe width (β) by the equation:
\[ \Delta = N \cdot \beta \]
where \( \beta \) is the fringe width.

### Step 3: Express fringe width in terms of wavelength
The fringe width can be expressed as:
\[ \beta = \frac{\lambda D}{d} \]
where \( \lambda \) is the wavelength of light, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

### Step 4: Substitute the expressions into the path difference equation
From the above, we can equate the two expressions for path difference:
\[ (n - 1) \cdot t = N \cdot \beta \]
Substituting for \( \beta \):
\[ (n - 1) \cdot t = N \cdot \frac{\lambda D}{d} \]

### Step 5: Solve for thickness (t)
Rearranging gives:
\[ t = \frac{N \cdot \lambda D}{(n - 1) \cdot d} \]
Since \( D \) and \( d \) will cancel out in the context of fringe displacement, we can simplify it to:
\[ t = \frac{N \cdot \lambda}{(n - 1)} \]

### Step 6: Substitute known values
Given:
- \( N = 30 \)
- \( \lambda = 6 \times 10^{-5} \) cm
- \( n = 1.5 \)

Substituting these values:
\[ t = \frac{30 \cdot (6 \times 10^{-5})}{(1.5 - 1)} \]
\[ t = \frac{30 \cdot (6 \times 10^{-5})}{0.5} \]
\[ t = 30 \cdot 12 \times 10^{-5} \]
\[ t = 360 \times 10^{-5} \text{ cm} \]
\[ t = 3.6 \times 10^{-3} \text{ cm} \]

### Final Answer
The thickness of the glass plate is:
\[ t = 3.6 \times 10^{-3} \text{ cm} \]