Question

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 ___$mm$

Hint:

The shortest distance between two coinciding bright fringes for two wavelengths can be found using the condition for fringe coincidence.

Answer 1

Correct Answer: 48

Step 1: The fringe width for the first wavelength (\( \lambda_1 \)) is given by:
\( \omega_1 = \frac{\lambda_1 D}{d} \)

\( \lambda_1 = 800 \, \text{nm} = 800 \times 10^{-9} \, \text{m} \), \( D = 7 \, \text{m} \), and \( d = 0.35 \, \text{mm} = 0.35 \times 10^{-3} \, \text{m} \)
So, \( \omega_1 = \frac{800 \times 10^{-9} \times 7}{0.35 \times 10^{-3}} = 16 \, \text{mm} \)

Step 2: The fringe width for the second wavelength (\( \lambda_2 \)) is given by:
\( \omega_2 = \frac{\lambda_2 D}{d} \)

\( \lambda_2 = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
So, \( \omega_2 = \frac{600 \times 10^{-9} \times 7}{0.35 \times 10^{-3}} = 12 \, \text{mm} \)

Step 3: The least common multiple (LCM) of \( \omega_1 \) and \( \omega_2 \) gives the distance where both fringes will coincide:
LCM(\( \omega_1, \omega_2 \)) = LCM(16, 12) = 48 mm