In Young's experiment the distance between two slits is `d/3` and the distance between the screen and the slits is 3D. The number of fringes in `1/3` metre on the screen, formed by monochromatic light of wavelength `3lambda`, will be:

Options

A. `d/(9Dlambda)`
B. `d/(27Dlambda)`
C. `d/(81Dlambda)`
D. `d/(Dlambda)`

Answer

To solve the problem, we need to find the number of fringes formed on the screen in Young's double slit experiment given the parameters.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Distance between the slits, \( d = \frac{d}{3} \)
- Distance from the slits to the screen, \( D = 3D \)
- Wavelength of light, \( \lambda = 3\lambda \)
- Length of the section on the screen, \( L = \frac{1}{3} \) m

2. **Calculate the Fringe Width (\( \beta \)):**
The fringe width in Young's experiment is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
Substituting the values:
\[ \beta = \frac{3\lambda \cdot 3D}{\frac{d}{3}} = \frac{9\lambda D}{\frac{d}{3}} = \frac{27\lambda D}{d} \]

3. **Calculate the Number of Fringes (\( N \)):**
The number of fringes that fit into a given length \( L \) on the screen is given by:
\[ N = \frac{L}{\beta} \]
Substituting \( L = \frac{1}{3} \) m and \( \beta = \frac{27\lambda D}{d} \):
\[ N = \frac{\frac{1}{3}}{\frac{27\lambda D}{d}} = \frac{d}{81\lambda D} \]

4. **Final Expression for the Number of Fringes:**
Thus, the number of fringes formed on the screen is:
\[ N = \frac{d}{81\lambda D} \]

### Conclusion:
The number of fringes in \( \frac{1}{3} \) metre on the screen is \( \frac{d}{81\lambda D} \).