Chapter 35, Problem 1DQ

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.