Question

In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by 5 × 10^–2 m towards the slits, the change in fringe width is 3 × 10^–3 cm. If the distance between the slits is 1 mm, then the wavelength of the light will be _______ nm.

Answer 1

Correct Answer: 600

To solve for the wavelength of light in a double slit experiment, we begin with the formula for fringe width:

β = ^λD⁄_d

where β is the fringe width, λ is the wavelength of light, D is the distance from the slits to the screen, and d is the slit separation.

The change in fringe width (Δβ) when the screen moves can be expressed as:

Δβ = ^λΔD⁄_d

We know:

• Δβ = 3 × 10^–3 cm = 3 × 10^–5 m
• ΔD = –5 × 10^–2 m (since the screen moves towards the slits)
• d = 1 mm = 1 × 10^–3 m

Substituting these values into the formula, we have:

3 × 10^–5 = ^{λ( –5 × 10–2 )}⁄_{1 × 10^–3}

Solving for λ gives:

λ = ^{3 × 10–5 × 1 × 10–3}⁄_{–5 × 10^–2} = –6 × 10^–7 m

Since the wavelength is always positive, λ = 6 × 10^–7 m.

Converting this to nm:

λ = 6 × 10^–7 m × 10⁹ nm/m = 600 nm

The wavelength of the light is 600 nm, which is within the expected range of 600,600 nm.