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 will use the relationship between the number of fringes, the wavelength of light, and the distance on the screen in the Young's Double Slit Experiment (YDSE).

### Step-by-Step Solution:

1. **Understanding the relationship**: In YDSE, the number of fringes (n) observed on a segment of the screen is given by the formula:
\[ n = \frac{x D}{\lambda} \]
where:
- \( n \) = number of fringes,
- \( x \) = width of the segment of the screen,
- \( D \) = distance between the slits and the screen,
- \( \lambda \) = wavelength of the light used.

2. **Setting up the equations**: For the first wavelength \( \lambda_1 = 500 \, \text{nm} \) with \( n_1 = 15 \) fringes, we can write:
\[ 15 = \frac{x D}{\lambda_1} \]
For the second wavelength \( \lambda_2 \) with \( n_2 = 10 \) fringes, we can write:
\[ 10 = \frac{x D}{\lambda_2} \]

3. **Equating the two expressions**: Since both equations represent the same segment of the screen, we can set them equal to each other:
\[ \frac{15}{\lambda_1} = \frac{10}{\lambda_2} \]

4. **Substituting the known values**: Substitute \( \lambda_1 = 500 \, \text{nm} \):
\[ \frac{15}{500} = \frac{10}{\lambda_2} \]

5. **Cross-multiplying to solve for \( \lambda_2 \)**:
\[ 15 \lambda_2 = 500 \times 10 \]
\[ 15 \lambda_2 = 5000 \]
\[ \lambda_2 = \frac{5000}{15} \]

6. **Calculating \( \lambda_2 \)**:
\[ \lambda_2 = \frac{5000}{15} = 333.33 \, \text{nm} \]

7. **Final answer**: The value of \( \lambda_2 \) is approximately \( 333.33 \, \text{nm} \).