Question

In Young's double slit experiment, the slits are 2mm apart and are illuminated by photons of two wavelengths $\lambda_1 = 12000 \mathring A $ and $\lambda_2 = 10000 \mathring A.$ At what minimum distance from the common central bright fringe on the screen 2m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other ?

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

Answer 1

The Correct Option is C

To solve this problem, we need to find the position on the screen where the bright fringe due to the wavelength $\lambda_1 = 12000 \mathring{A}$ coincides with the bright fringe due to the wavelength $\lambda_2 = 10000 \mathring{A}$.

In Young's double slit experiment, the position of bright fringes on the screen is given by the formula:

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

where:

• n is the fringe order (an integer representing the number of the fringe from the central maximum)
• λ is the wavelength of light used
• D is the distance from the slits to the screen (D = 2 \text{ m})
• d is the distance between the slits (d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m})

For the fringes of the two wavelengths to coincide, their positions must be equal, i.e.,

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

This simplifies to:

n_1 \lambda_1 = n_2 \lambda_2

where n_1 and n_2 are integers. Dividing through gives:

\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{10000}{12000} = \frac{5}{6}

To find the smallest solution, choose n_1 = 5 and n_2 = 6.

Using the position formula for one of the wavelengths, let's calculate the position on the screen where the fringes coincide:

x = \frac{n_1 \lambda_1 D}{d} = \frac{5 \times 12000 \times 10^{-10} \times 2}{2 \times 10^{-3}}

Calculating gives:

x = 6 \text{ mm}

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