Two light waves of wavelengths 800 and 600 nm are used in Young's double slit experiment to obtain interference fringes on a screen placed 7 m away from plane of slits. If the two slits are separated by 0.35 mm, then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be ______ mm.

Two light waves of wavelengths $$\lambda_1 = 800$$ nm and $$\lambda_2 = 600$$ nm are used in Young's double slit experiment. Screen is at $$D = 7$$ m and slit separation $$d = 0.35$$ mm.

$$ \beta_1 = \frac{\lambda_1 D}{d} = \frac{800 \times 10^{-9} \times 7}{0.35 \times 10^{-3}} = \frac{5600 \times 10^{-9}}{3.5 \times 10^{-4}} = 16 \times 10^{-3} \text{ m} = 16 \text{ mm} $$

$$ \beta_2 = \frac{\lambda_2 D}{d} = \frac{600 \times 10^{-9} \times 7}{0.35 \times 10^{-3}} = 12 \times 10^{-3} \text{ m} = 12 \text{ mm} $$

Bright fringes coincide when:

$$ n_1 \beta_1 = n_2 \beta_2 $$

$$ 16n_1 = 12n_2 \Rightarrow \frac{n_1}{n_2} = \frac{12}{16} = \frac{3}{4} $$

The smallest positive integers satisfying this ratio are $$n_1 = 3$$ and $$n_2 = 4$$.

$$ y = n_1 \beta_1 = 3 \times 16 = 48 \text{ mm} $$

Verification: $$n_2 \beta_2 = 4 \times 12 = 48$$ mm $$\checkmark$$

The shortest distance from the central bright maximum to the point of coincidence is 48 mm.