In Young's double slit experiment, fringes of width $\beta$ are produced on a screen kept at a distance of 1 m from the slit. When the screen is moved away by 5 x 10^-2 m, fringe width changes by 3 x 10^-5 m The separation between the slits is 1 x 10^-3 m The wavelength of the light used is ... nm

Explanation

In Young's double slit experiment, the fringe width β is given by the formula:

β = λD/d

Where:

• λ is the wavelength of the light,
• D is the distance between the slits and the screen,
• d is the separation between the slits.

Let the initial distance of the screen be D₁ = 1 m, and the fringe width is β₁ = λD₁/d.

After moving the screen by a distance of 5 × 10^-2 m, the new distance becomes D₂ = D₁ + 5 × 10^-2 = 1 + 5 × 10^-2 = 1.05 m.

The new fringe width is β₂ = λD₂/d.

The change in fringe width Δβ = β₂ - β₁ = 3 × 10^-5 m.

Substituting the expressions for β₁ and β₂, we get:

Δβ = λ(D2 - D1)/d

Substituting the given values:

3 × 10^-5 = λ(1.05 - 1)/1 × 10^-3

Simplifying:

3 × 10^-5 = λ × 50

Solving for λ:

λ = (3 × 10^-5) / 50 = 6 × 10^-7 m

λ = 600 nm

Final Answer

The wavelength of the light used is 600 nm.