Question

In YDSE experiment, the distance between 3rd maxima and 5th minima

• A. $\frac{\lambda d}{D}$
• B. $\frac{3 \lambda D}{2 d}$
• C. $\frac{dD}{\lambda}$
• D. $\frac{\lambda d}{4D}$

Answer 1

Answer: (B)

$\frac{3 \lambda D}{2 d}$

Answer 2

The problem asks for the distance between the 3rd maxima and 5th minima in a Young's Double Slit Experiment (YDSE).

The position of the n-th bright fringe (maxima) from the central maxima is given by: $y_n = \frac{n \lambda D}{d}$

For the 3rd maxima, $n=3$. $y_3 = \frac{3 \lambda D}{d}$

The position of the n-th dark fringe (minima) from the central maxima is given by: $y'_n = \frac{(2n - 1) \lambda D}{2d}$

For the 5th minima, $n=5$. $y'_5 = \frac{(2 \times 5 - 1) \lambda D}{2d} = \frac{(10 - 1) \lambda D}{2d} = \frac{9 \lambda D}{2d}$

To find the distance between the 3rd maxima and 5th minima, we take the absolute difference of their positions. Comparing $y_3$ and $y'_5$: $y_3 = \frac{3 \lambda D}{d} = \frac{6 \lambda D}{2d}$ $y'_5 = \frac{9 \lambda D}{2d}$ Since $y'_5 > y_3$, the 5th minima is further from the central maxima than the 3rd maxima.

The distance between them is: Distance = $y'_5 - y_3 = \frac{9 \lambda D}{2d} - \frac{6 \lambda D}{2d}$ Distance = $\frac{(9 - 6) \lambda D}{2d}$ Distance = $\frac{3 \lambda D}{2d}$