Question

In Young’s double slit experiment, slits are $ 3\, \text{mm} $ apart and illuminated with light of $ 3750\, \text{Å} $ and $ 7500\, \text{Å} $. Screen is placed $ 4\, \text{m} $ away. What is the minimum distance from the central fringe where bright fringes from both patterns coincide?

• A. 2 mm
• B. 3 mm
• C. 1 mm
• D. 8 mm

Hint:

Use \( y = \frac{n\lambda D}{d} \) to find fringe positions. Find least common fringe by LCM of wavelengths.

Answer 1

The Correct Option is C

To find where bright fringes coincide, calculate LCM of wavelengths: \[ \lambda_1 = 3750\, \text{Å}, \quad \lambda_2 = 7500\, \text{Å} \Rightarrow \text{LCM}(\lambda_1, \lambda_2) = 7500\, \text{Å} \] Path difference for bright fringe is: \[ \Delta y = \frac{n\lambda D}{d} \Rightarrow \text{minimum common fringe: } n = 2 \text{ for } \lambda = 3750,\quad n = 1 \text{ for } 7500 \] Let’s calculate: \[ y = \frac{\lambda D}{d} = \frac{7500 \times 10^{-10} \cdot 4}{3 \times 10^{-3}} = \frac{3 \times 10^{-6}}{3 \times 10^{-3}} = 1 \times 10^{-3} = 1\, \text{mm} \]