Question

If the Young's double slit experimental set up is immersed in a liquid of refractive index μ, the fringe width of the interference pattern observed is β. When the experiment is performed in air medium with the same experimental set up, the fringe width of the pattern will be

• A. β
• B. \(\frac{\beta}{\mu}\)
• C. (μ+1)β
• D. μβ
• E. (μ-1)β

Answer 1

The Correct Option is D

The fringe width \( \beta \) in Young's double slit experiment is given by: \[ \beta = \frac{\lambda D}{d} \] where \( \lambda \) is the wavelength of light, \( D \) is the distance between the slits and the screen, and \( d \) is the distance between the slits. When the experiment is performed in a medium with refractive index \( \mu \), the wavelength of light changes. The wavelength of light in a medium is given by: \[ \lambda_{\text{medium}} = \frac{\lambda}{\mu} \] Thus, the fringe width in the liquid becomes: \[ \beta_{\text{liquid}} = \frac{\lambda_{\text{medium}} D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu} \] When the experiment is returned to air (with refractive index \( \mu = 1 \)), the fringe width will be multiplied by \( \mu \).Therefore, the fringe width in air will be: \[ \mu \beta \]

Thus, the correct answer is (D): \( \mu \beta \).

Answer 2

The fringe width (\(\beta\)) in Young's double slit experiment is given by:

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

where:

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

When the setup is immersed in a liquid of refractive index \(\mu\), the wavelength of light changes to \(\lambda' = \frac{\lambda}{\mu}\). The new fringe width \(\beta'\) is given as \(\beta\).

\(\beta = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d}\)

When the experiment is performed in air (refractive index = 1), let the fringe width be \(\beta_{air}\). Then:

\(\beta_{air} = \frac{\lambda D}{d}\)

From the first equation, we have:

\(\beta = \frac{\beta_{air}}{\mu}\)

Therefore,

\(\beta_{air} = \mu \beta\)

So the fringe width in air is \(\mu\beta\).

Therefore, the answer is \(\mu\beta\).