In a Y.D.S.E, light of wavelength 500 nm is used. The separation between bright fringes is 7.5 mm., calculate the fringe width if light is replaced by another light of wavelength 650 nm and separation between the slits is increased to 1.5 times.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship for fringe width
In a Young's Double Slit Experiment (YDSE), the fringe width (β) is given by the formula:
\[ \beta = \frac{D \cdot \lambda}{d} \]
where:
- \(D\) is the distance from the slits to the screen,
- \(\lambda\) is the wavelength of the light,
- \(d\) is the separation between the slits.

### Step 2: Identify the initial conditions
From the problem, we know:
- Initial wavelength (\(\lambda_1\)) = 500 nm = \(500 \times 10^{-9}\) m
- Initial fringe width (\(\beta_1\)) = 7.5 mm = \(7.5 \times 10^{-3}\) m

Using the formula for fringe width, we can express it as:
\[ \beta_1 = \frac{D \cdot \lambda_1}{d} \]
Substituting the known values:
\[ 7.5 \times 10^{-3} = \frac{D \cdot (500 \times 10^{-9})}{d} \tag{1} \]

### Step 3: Set up the conditions for the second scenario
In the second scenario:
- New wavelength (\(\lambda_2\)) = 650 nm = \(650 \times 10^{-9}\) m
- The separation between the slits is increased to 1.5 times, so \(d_2 = 1.5d\).

The new fringe width (\(\beta_2\)) can be expressed as:
\[ \beta_2 = \frac{D \cdot \lambda_2}{d_2} = \frac{D \cdot \lambda_2}{1.5d} \tag{2} \]

### Step 4: Relate the two fringe widths
From equations (1) and (2), we can relate \(\beta_1\) and \(\beta_2\):
\[ \frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2} \cdot \frac{d_2}{d} \]
Substituting \(d_2 = 1.5d\):
\[ \frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2} \cdot 1.5 \]

### Step 5: Substitute known values
Substituting the known values:
\[ \frac{7.5 \times 10^{-3}}{\beta_2} = \frac{500 \times 10^{-9}}{650 \times 10^{-9}} \cdot 1.5 \]
Calculating the right side:
\[ \frac{500}{650} = \frac{5}{6.5} = \frac{10}{13} \]
Thus:
\[ \frac{7.5 \times 10^{-3}}{\beta_2} = \frac{10}{13} \cdot 1.5 \]
Calculating \( \frac{10}{13} \cdot 1.5 \):
\[ \frac{10 \cdot 1.5}{13} = \frac{15}{13} \]
So we have:
\[ \frac{7.5 \times 10^{-3}}{\beta_2} = \frac{15}{13} \]

### Step 6: Solve for \(\beta_2\)
Cross-multiplying gives:
\[ 7.5 \times 10^{-3} \cdot 13 = 15 \cdot \beta_2 \]
Calculating:
\[ 97.5 \times 10^{-3} = 15 \cdot \beta_2 \]
Thus:
\[ \beta_2 = \frac{97.5 \times 10^{-3}}{15} = 6.5 \times 10^{-3} \text{ m} = 6.5 \text{ mm} \]

### Final Answer
The fringe width when the light is replaced by another light of wavelength 650 nm and the separation between the slits is increased to 1.5 times is:
\[ \beta_2 = 6.5 \text{ mm} \]