In YDSE pattern with light of wavelength `lambda_1 = 500nm`, 15 fringes are obtained on a certain segment of screen. If number of fringes for light of wavelength `lambda_2` on same segment of screen is 10, then the value of `lambda_2` (in nm) is-

Answer

To solve the problem, we need to relate the number of fringes observed for two different wavelengths of light in a Young's Double Slit Experiment (YDSE).

### Step-by-Step Solution:

1. **Understand the Concept of Fringe Width**:
The fringe width (β) in a YDSE is given by the formula:
\[ \beta = \frac{\lambda \cdot D}{d} \]
where \( \lambda \) is the wavelength of the light, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

2. **Relate the Number of Fringes to Fringe Width**:
If we consider a segment of length \( y \) on the screen, the number of fringes \( n \) can be expressed as:
\[ y = n \cdot \beta \]
Thus, for two different wavelengths, we can write:
\[ y = n_1 \cdot \beta_1 = n_2 \cdot \beta_2 \]
where \( n_1 \) and \( n_2 \) are the number of fringes for wavelengths \( \lambda_1 \) and \( \lambda_2 \) respectively.

3. **Substituting for Fringe Width**:
For the first wavelength \( \lambda_1 \):
\[ y = n_1 \cdot \frac{\lambda_1 \cdot D}{d} \]
For the second wavelength \( \lambda_2 \):
\[ y = n_2 \cdot \frac{\lambda_2 \cdot D}{d} \]

4. **Equating the Two Expressions**:
Since \( y \) is the same for both cases, we can set the two equations equal to each other:
\[ n_1 \cdot \frac{\lambda_1 \cdot D}{d} = n_2 \cdot \frac{\lambda_2 \cdot D}{d} \]
The \( D \) and \( d \) cancel out:
\[ n_1 \cdot \lambda_1 = n_2 \cdot \lambda_2 \]

5. **Substituting Known Values**:
We know:
- \( n_1 = 15 \)
- \( \lambda_1 = 500 \, \text{nm} \)
- \( n_2 = 10 \)
We need to find \( \lambda_2 \):
\[ 15 \cdot 500 = 10 \cdot \lambda_2 \]

6. **Solving for \( \lambda_2 \)**:
\[ 7500 = 10 \cdot \lambda_2 \]
\[ \lambda_2 = \frac{7500}{10} = 750 \, \text{nm} \]

### Final Answer:
The value of \( \lambda_2 \) is **750 nm**.