Question

Two slits are made 0.1 mm apart and the screen is placed 2 m away. The fringe separation when a light of wavelength 500 nm is used is______.
Fill in the blank with the correct answer from the options given below.

• A. 1 cm
• B. 0.15 cm
• C. 0.1 cm
• D. 1.5 cm

Answer 1

The Correct Option is C

The problem involves understanding the interference pattern created by a double-slit experiment. The formula for fringe separation \( ( \Delta y ) \) in a double-slit experiment is given by:

\(\Delta y = \frac{\lambda L}{d}\)

where:

• \(\lambda\) is the wavelength of the light used
• \(L\) is the distance between the screen and the slits
• \(d\) is the distance between the slits

Given:

• \(\lambda = 500\ nm = 500 \times 10^{-9}\ m\)
• \(L = 2\ m\)
• \(d = 0.1\ mm = 0.1 \times 10^{-3}\ m\)

Substitute these values into the formula:

\(\Delta y = \frac{500 \times 10^{-9} \times 2}{0.1 \times 10^{-3}}\)

Calculate:

\(\Delta y = \frac{1000 \times 10^{-9}}{0.1 \times 10^{-3}} = \frac{1000 \times 10^{-9}}{0.0001}\)

\(\Delta y = 0.01\ m = 1\ cm = 0.1\ cm\)

Therefore, the fringe separation is 0.1 cm.

Answer 2

Let's analyze the fringe separation in Young's double-slit experiment.

1. Fringe Separation (β):

The fringe separation (β) is given by:

β = λD / d

Where:

• λ is the wavelength of light
• D is the distance from the slits to the screen
• d is the separation between the slits

2. Given Values:

• λ = 500 nm = 500 × 10^-9 m = 5 × 10^-7 m
• D = 2 m
• d = 0.1 mm = 0.1 × 10^-3 m = 1 × 10^-4 m

3. Calculate Fringe Separation (β):

β = (5 × 10^-7 m × 2 m) / (1 × 10^-4 m)

β = 10 × 10^-7 m / 10^-4 m

β = 10 × 10^-3 m

β = 1 × 10^-3 m

β = 1 mm

β = 0.1 cm

Therefore, the fringe separation is 0.1 cm.

The correct answer is:

Option 3: 0.1 cm