Question

In Young's double slit experiment, in an interference pattern, third minimum is observed exactly in front of one of the slits. If '\(d\)' is the distance between the slits and '\(D\)' is the distance between the slits and the screen, the wavelength of light used is

• A. \(d^{2}/3D\)
• B. \(d^{2}/4D\)
• C. \(d^{2}/5D\)
• D. \(d^{2}/6D\)

Hint:

Write the position of the third dark fringe and equate it to the distance of a slit from the central line, which is d/2.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In Young's experiment the n-th dark fringe lies at \(y = \dfrac{(2n-1)\lambda D}{2d}\) from the central bright fringe.

Step 2: Position of the third minimum
For \(n=3\), \(y = \dfrac{5\lambda D}{2d}\).

Step 3: Position in front of a slit
A point exactly in front of one slit is at a distance \(y = \dfrac{d}{2}\) from the central line.

Step 4: Calculation
Equate the two: \(\dfrac{d}{2} = \dfrac{5\lambda D}{2d}\).
So \(\lambda = \dfrac{d^{2}}{5D}\).

Final Answer:
The wavelength of the light is d squared divided by 5D. \[ \boxed{\text{(C) }d^{2}/5D} \]