Question

In a Young’s double-slit experiment, a beam of light consisting of two wavelengths 500 nm and 600 nm is used. The interference fringes are observed at a screen placed 1.8 m away from the plane of slits (slit separation 0.3 mm). Calculate the least distance from the central maximum where the bright fringes due to both the wavelengths coincide.

Hint:

For two wavelengths in YDSE: First find LCM condition \(n_1\lambda_1 = n_2\lambda_2\), then substitute in fringe formula.

Answer 1

Concept: In Young’s double slit experiment, position of bright fringe: \[ y = \frac{n \lambda D}{d} \] For coincidence of bright fringes: \[ n_1 \lambda_1 = n_2 \lambda_2 \]

Step 1: Given data
\[ \lambda_1 = 500 \text{ nm}, \quad \lambda_2 = 600 \text{ nm} \]

Step 2: Condition for coincidence
\[ n_1 \cdot 500 = n_2 \cdot 600 \] Divide: \[ \frac{n_1}{n_2} = \frac{600}{500} = \frac{6}{5} \] Smallest integers: \[ n_1 = 6, \quad n_2 = 5 \]

Step 3: Find position of coincidence
Using: \[ y = \frac{n \lambda D}{d} \] Take \( \lambda_1 \): \[ y = \frac{6 \times 500 \times 10^{-9} \times 1.8}{0.3 \times 10^{-3}} \]

Step 4: Simplification
\[ y = \frac{6 \times 500 \times 1.8}{0.3} \times 10^{-6} \] \[ = \frac{5400}{0.3} \times 10^{-6} = 18000 \times 10^{-6} \] \[ y = 1.8 \times 10^{-2} \text{ m} \] \[ y = 18 \text{ mm} \] Final Answer: \[ \boxed{18 \ \text{mm}} \]