Question

In Young's double slit experiment, in an interference pattern, a minimum is observed exactly in front of one slit. The distance between the two coherent sources is $d$ and $D$ is the distance between the source and screen. The possible wavelengths used are inversely proportional to

• A. $D, 5D, 9D, \dots$
• B. $D, 3D, 5D, \dots$
• C. $3D, 4D, 5D, \dots$
• D. $3D, 7D, 10D, \dots$

Hint:

Whenever you see a question involving a "minimum" (dark fringe) in an interference setup, look for odd integer sequences ($1, 3, 5, \dots$) in the solutions. Destructive interference is inherently tied to half-wavelength path differences, which naturally maps to odd multiples.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
In a classic Young's Double Slit Experiment (YDSE) setup, the two slits are separated by a small distance $d$, and the viewing screen is placed at a distance $D$ away.
A dark fringe (interference minimum) is observed on the screen at a position located exactly straight ahead from one of the slits. We need to find the sequence of parameters that the possible wavelengths $\lambda$ are inversely proportional to.

Step 2: Key Formula or Approach:
1. The central bright fringe forms along the perpendicular bisector of the two slits. The position directly in front of one slit is at a distance $y$ from this central maximum:
$$y = \frac{d}{2}$$ 2. The condition for observing a destructive interference minimum at a distance $y$ from the central bright fringe is given by:
$$y = (2n - 1)\frac{\lambda D}{2d} \quad \text{where } n = 1, 2, 3, \dots$$

Step 3: Detailed Explanation:
Equate the position of the slit line ($y = \frac{d}{2}$) to the general mathematical condition for a minimum:
$$\frac{d}{2} = (2n - 1)\frac{\lambda D}{2d}$$ Cancel out the factor of 2 from the denominators on both sides:
$$d = (2n - 1)\frac{\lambda D}{d}$$ Rearrange the equation to express the wavelength $\lambda$ explicitly:
$$d^2 = (2n - 1)\lambda D \implies \lambda = \frac{d^2}{(2n - 1)D}$$ For consecutive integer values of $n = 1, 2, 3, \dots$, the odd coefficient term $(2n - 1)$ generates the sequence $1, 3, 5, \dots$
Let's list the corresponding possible wavelengths:
$$\lambda_1 = \frac{d^2}{1 \cdot D}, \quad \lambda_2 = \frac{d^2}{3D}, \quad \lambda_3 = \frac{d^2}{5D}, \dots$$ This sequence shows that the possible wavelengths can be written generally as:
$$\lambda \propto \frac{1}{(2n - 1)D}$$ Therefore, the possible wavelengths used are inversely proportional to the sequence $D, 3D, 5D, \dots$

Step 4: Final Answer:
The wavelengths are inversely proportional to $D, 3D, 5D, \dots$, which matches option (B).