Question

In a Young’s double slit experiment, a laser light of wavelength 560 nm produces an interference pattern with consecutive bright fringes’ separation of 7.2 mm. Now another light is used to produce an interference pattern with consecutive bright fringes’ separation of 8.1 mm. The wavelength of the second light is:

• A. \( 680 \, \text{nm} \)
• B. \( 630 \, \text{nm} \)
• C. \( 650 \, \text{nm} \)
• D. \( 540 \, \text{nm} \)

Hint:

In interference experiments, the fringe separation is directly proportional to the wavelength of the light used. By using the ratio of fringe separations, we can easily find the unknown wavelength.

Answer 1

The Correct Option is B

In Young's double slit experiment, the fringe separation \( \Delta y \) is given by the formula: \[ \Delta y = \frac{\lambda L}{d} \] where: 
- \( \lambda \) is the wavelength of light, 
- \( L \) is the distance between the screen and the slits, 
- \( d \) is the separation between the slits. 
Let the wavelength of the first light be \( \lambda_1 = 560 \, \text{nm} \), and the fringe separation is \( \Delta y_1 = 7.2 \, \text{mm} \). The wavelength of the second light is \( \lambda_2 \), and the fringe separation is \( \Delta y_2 = 8.1 \, \text{mm} \). Since the experimental setup remains the same, we can equate the ratios of the fringe separations and wavelengths: \[ \frac{\Delta y_1}{\Delta y_2} = \frac{\lambda_1}{\lambda_2} \] Substituting the known values: \[ \frac{7.2}{8.1} = \frac{560}{\lambda_2} \] Solving for \( \lambda_2 \): \[ \lambda_2 = \frac{560 \times 8.1}{7.2} \approx 630 \, \text{nm} \] 
Conclusion: The wavelength of the second light is \( 630 \, \text{nm} \). 
Final Answer: \[ \boxed{630 \, \text{nm}} \]