Chapter 35, Problem 35.1DQ

A two-slit interference experiment is set up, and the fringes are displayed on a screen. Then the whole apparatus is immersed in the nearest swimming pool. How does the fringe pattern change?

To determine

The explanation for the change in the fringe pattern when the whole apparatus of the two-slit interference experiment set up is immersed in the nearest swimming pool.

Explanation of Solution

Formula to calculate position of $m\text{th}$ bright fringe in air medium,

$y_m=R\frac{m{\lambda }_{\text{o}}}{d}$ (I)

• $R$ is the distance from slits to screen
• $m$ is the order of the fringe
• ${\lambda }_{\text{o}}$ is the wavelength of the light in air medium
• $d$ is the distance between slits

Formula to calculate position of $m\text{th}$ bright fringe in water medium,

$y_m\text{'}=R\frac{m{\lambda }_{\text{w}}}{d}$ (II)

• $R$ is the distance from slits to screen
• $m$ is the order of the fringe
• ${\lambda }_{\text{w}}$ is the wavelength of the light in water medium
• $d$ is the distance between slits

Divide equation (II) with equation (I),

$\begin{array}{c}\frac{y_m\text{'}}{y_m}=\frac{\left(R\frac{m{\lambda }_{\text{w}}}{d}\right)}{\left(R\frac{m{\lambda }_{\text{o}}}{d}\right)}\\ =\frac{{\lambda }_{\text{w}}}{{\lambda }_{\text{o}}}\end{array}$

Formula to calculate the wavelength of the light in water medium,

${\lambda }_{\text{w}}=\frac{{\lambda }_{\text{o}}}{n}$

• $n$ is the refractive index of the water

Substitute $\frac{{\lambda }_{\text{o}}}{n}$ for ${\lambda }_{\text{w}}$ to find $y_m\text{'}$,

$\begin{array}{c}\frac{y_m\text{'}}{y_m}=\frac{\left(\frac{{\lambda }_{\text{o}}}{n}\right)}{{\lambda }_{\text{o}}}\\ =\frac{1}{n}\end{array}$

Substitute $1.33$ for $n$ to find $y_m\text{'}$,

$\begin{array}{c}\frac{y_m\text{'}}{y_m}=\frac{1}{1.33}\\ =\frac{3}{4}\\ y_m\text{'}=\frac{3y_m}{4}\end{array}$

Conclusion:

Therefore, the new fringe pattern becomes $\frac{3}{4}$ times the old fringe pattern as the whole apparatus is immersed in water.