Question

In Young's double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths, $λ_1$ =12000 Å and $λ_2$= 10000 Å. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference patter coincide with a bright fringe from the other?

• A. 8 mm
• B. 6 mm
• C. 4 mm
• D. 3 mm

Answer 1

The Correct Option is B

To determine the minimum distance where a bright fringe from one interference pattern coincides with a bright fringe from the other in Young's double slit experiment, we must find a common position where bright fringes of both wavelengths overlap.

Step 1: Understand the formula for bright fringes

In Young's double slit experiment, the position of the n_th bright fringe is given by:

y_n = \frac{n \lambda D}{d}

• y_n is the distance of the n_th bright fringe from the central maximum.
• \lambda is the wavelength of light.
• D is the distance from the slits to the screen.
• d is the separation between the slits.

Step 2: Equate the conditions for coincidence

Bright fringes will coincide when:

\frac{n_1 \lambda_1}{d} = \frac{n_2 \lambda_2}{d}

This simplifies to:

n_1 \lambda_1 = n_2 \lambda_2

Here, \lambda_1 = 12000 \, \text{Å} and \lambda_2 = 10000 \, \text{Å}. Thus,

n_1 \cdot 12000 = n_2 \cdot 10000

Dividing through by 1000, we get:

12n_1 = 10n_2

This implies:

\frac{n_1}{n_2} = \frac{5}{6}

The minimum values of n_1 and n_2 that satisfy this ratio are n_1 = 5 and n_2 = 6.

Step 3: Calculate the position of the coinciding fringe

Using the bright fringe position formula for either wavelength (let's use \lambda_1):

y = \frac{5 \cdot 12000 \cdot 2}{2000}

Simplifying:

y = \frac{600000}{2000} = 6 \, \text{mm}

Therefore, the minimum distance from the central bright fringe where a bright fringe from one pattern coincides with a bright fringe from the other is 6 mm.