Question

In a Young' double slit experiment. if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference.

• A. $5\frac{\lambda}{2}$
• B. $10\frac{\lambda}{2}$
• C. $9\frac{\lambda}{2}$
• D. $11\frac{\lambda}{2}$

Answer 1

The Correct Option is C

To solve this problem, we need to understand the concept of minima in the Young's double slit experiment. In this experiment, the condition for destructive interference (minima) is given by the path difference between light from the two slits being an odd multiple of half-wavelengths.

Concept Explanation

In Young's double slit experiment, the condition for minima is given by the formula:

d \sin \theta = \left(n + \frac{1}{2} \right)\lambda

where:

• d is the distance between the slits.
• \theta is the angle of diffraction.
• n is the order of the minimum (0, 1, 2,...).
• \lambda is the wavelength of the light.

Calculation for the Fifth Minimum

1. We are asked to find the path difference for the fifth minimum. For minima, the path difference is given by: \Delta = \left(n + \frac{1}{2} \right)\lambda.
2. The order of the minimum is n = 4 (since counting starts from 0).
3. The path difference will be: \Delta = \left(4 + \frac{1}{2}\right)\lambda = \frac{9\lambda}{2}.

Conclusion

Therefore, the path difference corresponding to the fifth minimum is 9\frac{\lambda}{2}. Hence, the correct answer is:

Option: 9\frac{\lambda}{2}