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 requires determining the fringe separation \( (\Delta y) \) in a double-slit experiment. The formula is:

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

where:

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

Given values:

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

Substituting the values:

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

Calculation:

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

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

The fringe separation is 0.1 cm.