15. In a two slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If screen is moved by 5 × 10–2 m towards the slits, then change in fringe width is 3 × 10–5 m. If the distance between slits is 10–3 m then wavelength of the light used will be (a) 4000 Å (b) 6000 Å (c) 5890 Å (d) 8000 Å

To determine the wavelength of light used in a double-slit experiment, we can use the relationship between fringe width, distance to the screen, and the wavelength itself. The fringe width (β) is given by the formula:

Understanding Fringe Width

The fringe width can be expressed as:

• β = λL / d

Where:

• β = fringe width
• λ = wavelength of the light
• L = distance from the slits to the screen
• d = distance between the slits

Analyzing the Problem

In this scenario, we know the following:

• The distance between the slits (d) = 10^-3 m
• The change in fringe width (Δβ) = 3 × 10^-5 m
• The screen is moved closer by 5 × 10^-2 m.

Let’s denote the initial distance from the slits to the screen as L. When the screen is moved closer, the new distance becomes (L - 5 × 10^-2 m).

Calculating the Change in Fringe Width

The initial fringe width (β₁) can be expressed as:

• β₁ = λL / d

The new fringe width (β₂) after moving the screen is:

• β₂ = λ(L - 5 × 10^-2) / d

The change in fringe width (Δβ) is given by:

• Δβ = β₂ - β₁

Substituting the expressions for β₁ and β₂ gives:

• Δβ = λ(L - 5 × 10^-2) / d - λL / d

This simplifies to:

• Δβ = λ(-5 × 10^-2) / d

Solving for Wavelength

Now, we can rearrange this equation to solve for λ:

• λ = Δβ * d / (-5 × 10^-2)

Substituting the known values:

• Δβ = 3 × 10^-5 m
• d = 10^-3 m

We get:

• λ = (3 × 10^-5 m * 10^-3 m) / (-5 × 10^-2 m)

Calculating this gives:

• λ = (3 × 10^-8 m) / (-5 × 10^-2)

λ = -6 × 10^-7 m = 6000 Å (since 1 m = 10¹⁰ Å).

Final Answer

Thus, the wavelength of the light used in the experiment is 6000 Å, which corresponds to option (b).