A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10 th bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same 10th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be _________ mm .

In Young's double-slit interference experiment, the position of the nth bright fringe from the central maximum is given by the formula:

$$ x = \frac{n \lambda D}{d} $$

where:
- $$n$$ is the fringe order,
- $$\lambda$$ is the wavelength of light,
- $$D$$ is the distance from the slits to the screen,
- $$d$$ is the separation between the slits.

For the first light source:
Wavelength, $$\lambda_1 = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$$,
Position of the 10th bright fringe, $$x_1 = 10 \text{ mm} = 0.01 \text{ m}$$,
Fringe order, $$n = 10$$.

Substituting into the formula:

$$ x_1 = \frac{10 \lambda_1 D}{d} = 0.01 \quad ...(1) $$

For the second light source:
Wavelength, $$\lambda_2 = 660 \text{ nm} = 660 \times 10^{-9} \text{ m}$$,
Fringe order, $$n = 10$$ (same as before),
We need to find the new position $$x_2$$.

The formula becomes:

$$ x_2 = \frac{10 \lambda_2 D}{d} \quad ...(2) $$

Since the experimental setup ($$D$$ and $$d$$) remains unchanged, we can take the ratio of equations (2) and (1):

$$ \frac{x_2}{x_1} = \frac{\frac{10 \lambda_2 D}{d}}{\frac{10 \lambda_1 D}{d}} = \frac{\lambda_2}{\lambda_1} $$

Therefore:

$$ x_2 = x_1 \cdot \frac{\lambda_2}{\lambda_1} $$

Substituting the given values (keeping units consistent in mm and nm for simplicity):
$$x_1 = 10 \text{ mm}$$,
$$\lambda_1 = 600 \text{ nm}$$,
$$\lambda_2 = 660 \text{ nm}$$.

So:

$$ x_2 = 10 \times \frac{660}{600} $$

Simplifying the fraction:

$$ \frac{660}{600} = \frac{66}{60} = \frac{11}{10} = 1.1 $$

Thus:

$$ x_2 = 10 \times 1.1 = 11 \text{ mm} $$

The distance of the 10th bright fringe from the central maximum for the second wavelength is 11 mm.