Question

In Young’s double slit experiment, when wavelength used is 600 Å and the screen is 40 cm from the slits, then the fringes are 0.012 cm apart. What is the distance between the slits?

• A. 0.024 cm
• B. 2.4 cm
• C. 0.24 cm
• D. 0.2 cm

Hint:

Fringe width $β = \fracλ Dd$.

Answer 1

The Correct Option is D

To find the distance between the slits in Young's double-slit experiment, we use the formula for fringe separation (or fringe width) in a double-slit experiment:

\(w = \frac{\lambda \times D}{d}\)

Where,

• \(w\) is the fringe separation,
• \(\lambda\) is the wavelength of light used,
• \(D\) is the distance between the slits and the screen,
• \(d\) is the distance between the slits.

We are given:

• Wavelength, \(\lambda = 600 \, \text{\AA} = 600 \times 10^{-10} \, \text{cm}\) (since 1 Å = 10^-10 m),
• Distance to the screen, \(D = 40 \, \text{cm}\),
• Fringe separation, \(w = 0.012 \, \text{cm}\).

We need to find the distance between the slits \(d\).

Rearrange the formula to solve for \(d\):

\(d = \frac{\lambda \times D}{w}\)

Substitute the given values:

\(d = \frac{600 \times 10^{-10} \, \text{cm} \times 40 \, \text{cm}}{0.012 \, \text{cm}}\)

Calculate the above expression:

\(d = \frac{24000 \times 10^{-10}}{0.012}\)

This simplifies to:

\(d = 2 \times 10^{-1} \, \text{cm} = 0.2 \, \text{cm}\)

Therefore, the distance between the slits is 0.2 cm. This matches the correct answer.