Question

A double-slit arrangement produces bright interference fringes for sodium light (a distinct yellow light at a wavelength of $\left(\lambda =589\mathrm{mm}\right)$. The fringes are angularly separated by $\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{°}$ near the centre of the pattern. What is the angular fringe separation if the entire arrangement is immersed in water, which has an index of refraction of $\mathbf{1}\mathbf{.}\mathbf{33}$?

Short answer

The angular fringe separation is $0.23°$

Step-by-step solution

Given data

Wavelength of light$\lambda =589 \text{nm}$

The reactive index of water $n=\text{1.33}$

Angularly separation of fringenear the centre of the pattern$\Delta \theta =0.30°$

Definition and concept used of interference fringe

Interference fringe, a bright or dark band caused by beams of light that are in phase or out of phase with one another.

In double slit experiment the expression for the angular separation $\Delta \theta$between the fringes and wavelength$\lambda$ of light used is

$\Delta \theta =\frac{\lambda }{d}$

Here d is the distance between the two slits.

$\left(n=1.33\right)$

Now, we have to find the wavelength of light in the water refractive index

In the double sheet experiment wavelength of light $\text{'}\lambda \text{'}=589\text{mm}$

For the bright fringe,

The angle $\Delta \theta =0.30°$

The reactive index of water $\left(n=\text{1.33}\right)$

The angular separation between the fringes when we observed the interference pattern in air medium

Determine the angular fringe separation

The angular fringe separation

$\Delta \theta =\frac{\lambda }{d}$ ...(i)

The angular separation between the fringes when we observed the interference pattern in water medium

$\Delta {\theta }_w=\frac{{\lambda }_w}{d}$ ...(ii)

Here ${\lambda }_w$ is wave length of light in the water medium

${\lambda }_w=\frac{\lambda }{n}$ ...(iii)

Substitution equation (iii) in equation (ii) we get

$\Delta {\theta }_w=\frac{\lambda }{nd}$ ...(iv)

Dividing equation (iv) and (i), we get

$\frac{\Delta {\theta }_w}{\Delta \theta }=\frac{1}{n}$

And rearranging the above equation and substitute all values, we get angular separation between the fringes as

$\begin{array}{rcl}\Delta {\theta }_w& =& \frac{1}{n}\left(\Delta \theta \right)\\ & =& \frac{1}{1.33}\left(0.30°\right)\\ & =& 0.23°\end{array}$

Hence, the wavelength of light in the water refractive index is $0.23°$