Answer to Question #298890 in Optics for Chemistry

Question

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?

Answer 1

Position of $m^{th}$ bright fringe in air medium,

$y_m=R\frac{m\lambda_\omicron}{d}$..... (1)

Position of $m^{th}$ bright fringe in water medium,

${y_m}'=R\frac{m\lambda_W}{d}$..... (2)

in equation (1) and (2)

R is the distance from slits to screen

m is the order of the fringe

$\lambda_W$ is the wavelength of the light in water medium

${\lambda_\omicron}$ is the wavelength of the light in air medium

d is the distance between slits

Divide equation (II) with equation (I),

$\frac{{y_m}'}{{y_m}}=\frac{R\frac{m\lambda_W}{d}}{R\frac{m\lambda_\omicron}{d}}$

$=\frac{\lambda_W}{\lambda_\omicron}$

The wavelength of the light in water medium is given by,

$\lambda_W=\frac{\lambda_\omicron}{n}$

where refractive index of the water is n,

Substitute $\frac{\lambda_\omicron}{n}$ for $\lambda_W$ to find ${y_m}'$ ,

$\frac{{y_m}'}{{y_m}}=\frac{\frac{\lambda_\omicron}{n}}{\lambda_\omicron}$

$=\frac{1}{n}$

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

$\frac{{y_m}'}{{y_m}}=\frac{1}{1.33}$

$=\frac{3}{4}$

${y_m}'=\frac{3{y_m}}{4}$

Therefore, the new fringe pattern is $\frac{3}{4}$ times the old fringe pattern.