Question

In Young's experiment, light of wavelength 6000Å is used to produce fringes of width 0.8 mm at a distance of 2.5 m. If the whole experiment is deep in a liquid of refractive index 1.6, then fringe width will be :

• A. 0.5 mm
• B. 0.6 mm
• C. 0.4 mm
• D. 0.2 mm

Answer 1

Answer: (A)

0.5 mm

Answer 2

The fringe width ($\beta$) in Young's Double Slit Experiment is directly proportional to the wavelength ($\lambda$) of light used. The formula for fringe width in air is given by:

$\beta_{air} = \frac{\lambda_{air} D}{d}$

where $\lambda_{air}$ is the wavelength of light in air, $D$ is the distance of the screen from the slits, and $d$ is the separation between the slits.

When the entire experiment is immersed in a liquid of refractive index $\mu$, the wavelength of light changes. The new wavelength in the liquid ($\lambda_{liquid}$) is related to the wavelength in air by:

$\lambda_{liquid} = \frac{\lambda_{air}}{\mu}$

The distance $D$ and slit separation $d$ remain unchanged. Therefore, the new fringe width in the liquid ($\beta_{liquid}$) will be:

$\beta_{liquid} = \frac{\lambda_{liquid} D}{d}$

Substitute the expression for $\lambda_{liquid}$:

$\beta_{liquid} = \frac{(\frac{\lambda_{air}}{\mu}) D}{d}$

$\beta_{liquid} = \frac{1}{\mu} \left(\frac{\lambda_{air} D}{d}\right)$

Recognizing that $\frac{\lambda_{air} D}{d}$ is $\beta_{air}$, we get:

$\beta_{liquid} = \frac{\beta_{air}}{\mu}$

Given values:

Fringe width in air, $\beta_{air} = 0.8 \, mm$ Refractive index of the liquid, $\mu = 1.6$

Now, substitute these values into the formula:

$\beta_{liquid} = \frac{0.8 \, mm}{1.6}$

$\beta_{liquid} = 0.5 \, mm$

The fringe width in the liquid will be 0.5 mm.