Question

In Young's double slit experiment, the slits are \(2\ \text{mm}\) apart and are illuminated by photons of two wavelengths \(\lambda _{1}=6000\ \text{Å}\) and \(\lambda _{2}=5000\ \text{Å}\) one after the other. At what minimum distance from the common central bright fringe on the screen \(2\ \text{m}\) from the slit will, a bright fringe from one interference pattern coincide with a bright fringe from the other?

• A. \(8\ \text{mm}\)
• B. \(6\ \text{mm}\)
• C. \(4\ \text{mm}\)
• D. \(3\ \text{mm}\)

Hint:

Find each fringe width, then look for the smallest distance that is a whole-number multiple of both widths.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The fringe width in Young's experiment is \(\beta = \dfrac{\lambda D}{d}\). Bright fringes of the two patterns coincide where the distance from the centre is a whole multiple of both fringe widths.

Step 2: Calculation of fringe widths
\(\beta_1 = \dfrac{6000\times 10^{-10}\times 2}{2\times 10^{-3}} = 0.6\ \text{mm}\)
\(\beta_2 = \dfrac{5000\times 10^{-10}\times 2}{2\times 10^{-3}} = 0.5\ \text{mm}\)

Step 3: Condition for coincidence
We need \(n_1\beta_1 = n_2\beta_2\), so \(0.6\,n_1 = 0.5\,n_2\), that is \(\dfrac{n_1}{n_2}=\dfrac{5}{6}\). The smallest integers are \(n_1 = 5\) and \(n_2 = 6\).

Step 4: Minimum distance
\(y = 5\times 0.6 = 6\times 0.5 = 3\ \text{mm}\).

Final Answer:
The first coincidence of bright fringes is at 3 mm from the central fringe. \[ \boxed{\text{(D) }3\ \text{mm}} \]