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 7.2 mm .calculate the wavelength of the second light .

Answer

To solve the problem, we need to use the relationship between fringe width and wavelength in an interference pattern. The fringe width (β) is given by the formula:

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

where:
- \( \beta \) is the fringe width,
- \( \lambda \) is the wavelength of the light,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.

Since \( D \) and \( d \) are constants for both lasers, we can say that the fringe width is directly proportional to the wavelength:

\[ \beta_1 \propto \lambda_1 \quad \text{and} \quad \beta_2 \propto \lambda_2 \]

From this, we can establish the ratio:

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

Given:
- \( \lambda_1 = 630 \, \text{nm} \)
- \( \beta_1 = 8.1 \, \text{mm} \)
- \( \beta_2 = 7.2 \, \text{mm} \)

We need to find \( \lambda_2 \). Rearranging the ratio gives us:

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

Now, substituting the values:

\[ \lambda_2 = \frac{7.2 \, \text{mm}}{8.1 \, \text{mm}} \cdot 630 \, \text{nm} \]

Calculating the fraction:

\[ \frac{7.2}{8.1} = 0.8889 \]

Now, substituting this back into the equation for \( \lambda_2 \):

\[ \lambda_2 = 0.8889 \cdot 630 \, \text{nm} \approx 560 \, \text{nm} \]

Thus, the wavelength of the second light is approximately \( 560 \, \text{nm} \).

### Summary of the Steps:
1. Write down the relationship between fringe width and wavelength.
2. Establish the proportionality between the two sets of data.
3. Rearrange the equation to solve for the unknown wavelength.
4. Substitute the known values and calculate the result.