A source emitting light of wavelengths 480 nm and 600 nm is used in a double slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.

Explanation

To solve the problem of finding the linear separation between the first maximum (next to the central maximum) corresponding to two wavelengths in a double slit interference experiment, we can follow these steps:

### Step 1: Understand the parameters given
- Wavelengths: \( \lambda_1 = 480 \, \text{nm} = 480 \times 10^{-9} \, \text{m} \)
- Wavelengths: \( \lambda_2 = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Slit separation: \( d = 0.25 \, \text{mm} = 0.25 \times 10^{-3} \, \text{m} \)
- Distance to the screen: \( D = 150 \, \text{cm} = 1.5 \, \text{m} \)

### Step 2: Calculate the fringe width for each wavelength
The fringe width (\( \beta \)) in a double slit experiment is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]

#### For \( \lambda_1 = 480 \, \text{nm} \):
\[ \beta_1 = \frac{480 \times 10^{-9} \, \text{m} \times 1.5 \, \text{m}}{0.25 \times 10^{-3} \, \text{m}} \]
Calculating this gives:
\[ \beta_1 = \frac{480 \times 1.5}{0.25} \times 10^{-6} = \frac{720}{0.25} \times 10^{-6} = 2880 \times 10^{-6} \, \text{m} = 2.88 \, \text{mm} \]

#### For \( \lambda_2 = 600 \, \text{nm} \):
\[ \beta_2 = \frac{600 \times 10^{-9} \, \text{m} \times 1.5 \, \text{m}}{0.25 \times 10^{-3} \, \text{m}} \]
Calculating this gives:
\[ \beta_2 = \frac{600 \times 1.5}{0.25} \times 10^{-6} = \frac{900}{0.25} \times 10^{-6} = 3600 \times 10^{-6} \, \text{m} = 3.6 \, \text{mm} \]

### Step 3: Find the linear separation between the first maxima
The linear separation between the first maximum corresponding to the two wavelengths is given by:
\[ \Delta y = \beta_2 - \beta_1 \]
Substituting the values we calculated:
\[ \Delta y = 3.6 \, \text{mm} - 2.88 \, \text{mm} = 0.72 \, \text{mm} \]

### Final Answer
The linear separation between the first maximum corresponding to the two wavelengths is \( 0.72 \, \text{mm} \).

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