Laser light of wavelength 630 nm incident on a pair of slits produces an interference pattern in which the fringes are seprated by 8.1mm .A second laser light produces an interference pattern in which the fringes are seprated by 3.6 mm.Calculate the wavelength of the second light

Answer

To solve the problem, we need to use the relationship between the fringe separation (β) and the wavelength (λ) of the light in an interference pattern. The formula that relates these quantities is given by:

\[ \beta = \frac{\lambda D}{d} \]

Where:
- \(\beta\) = fringe separation
- \(\lambda\) = wavelength of the light
- \(D\) = distance from the slits to the screen
- \(d\) = distance between the slits

Since \(D\) and \(d\) are constant for both light sources, we can conclude that the fringe separation is directly proportional to the wavelength of the light. Therefore, we can write:

\[ \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} \]

### Step-by-step Solution:

1. **Identify the given values:**
- Wavelength of the first light (\(\lambda_1\)) = 630 nm
- Fringe separation for the first light (\(\beta_1\)) = 8.1 mm
- Fringe separation for the second light (\(\beta_2\)) = 3.6 mm
- Wavelength of the second light (\(\lambda_2\)) = ?

2. **Set up the proportion:**
Using the relationship derived above:
\[ \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} \]

3. **Substitute the known values into the equation:**
\[ \frac{3.6 \text{ mm}}{8.1 \text{ mm}} = \frac{\lambda_2}{630 \text{ nm}} \]

4. **Cross-multiply to solve for \(\lambda_2\):**
\[ \lambda_2 = \frac{3.6 \text{ mm}}{8.1 \text{ mm}} \times 630 \text{ nm} \]

5. **Calculate the fraction:**
\[ \frac{3.6}{8.1} = 0.4444 \quad (\text{approximately}) \]

6. **Now calculate \(\lambda_2\):**
\[ \lambda_2 = 0.4444 \times 630 \text{ nm} \approx 280 \text{ nm} \]

### Final Answer:
The wavelength of the second light (\(\lambda_2\)) is approximately **280 nm**.