Question

In a Young's double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by

• A. 12
• B. 18
• C. 24
• D. 30

Hint:

 In Young's Double Slit Experiment, alternate light and dark bands, known as fringes, are observed on the screen.

Answer 1

The Correct Option is B

Fringe width, \(\omega=\frac{\lambda D}{d}∝\lambda\) 

When the wavelength is decreased from 600 nm to 400 nm, the fringe width will also decrease by a factor of \(\frac{4}{6}\) or \(\frac{2}{3}\) or the number of fringes in the same segment will increase by a factor of 3/2. 

Therefore, the number of fringes observed in the same segment \(=12\times\frac{3}{2}=18\) 

Note: Since \(\omega∝\lambda\), therefore, if the YDSE apparatus is immersed in a liquid of refractive index \(\mu\), the wavelength \(\lambda\) and thus the fringe width will decrease \(\mu\) times.

Answer 2

The wavelength of light is inversely proportional to the number of observed fringes.

Here, we use \(n_{1}​λ_{1}​=n_{2}​λ_{2}\)

\(⇒n_{2}​=\frac{n_{1​}λ_{1}}{λ_{2}}\)

\(n_{2}=\frac{12\times600}{400}=18\)

Answer 3

Monochromatic light falls on 2 slits that act as two coherent sources. In Young's Double Slit Experiment, alternate light and dark bands, known as fringes, are observed on the screen.

• The separation between 2 consecutive light or dark fringes is known as the fringe width.
• All fringes are of equal width in the experiment.

\(\beta = {D \lambda \over d}\)

• D is the distance between the screen and the slits
• d is the distance between slits
• \(\lambda\) is the wavelength of monochromatic light emitted via source

Calculation:

\(n\beta = {D \lambda \over d}\)

\(n_1\beta_1 = n_2\beta_2\)

n₂ = 12 x \( {600 \over 400}\)

= 18

Hence, if the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen is given by 18.