Question

A double slit arrangement produces interference fringes for sodium light ($\lambda =5890\mathring{A}$) that are ${0.40}^0$ apart. What is the angular fringe separation if the entire arrangement is immersed in water?

• A. ${0.10}^o$
• B. ${0.20}^o$
• C. ${0.30}^o$
• D. ${0.40}^o$

Answer 1

The correct option is C ${0.30}^o$

We know the distance between two slits $d=\frac{\lambda }{\sin \theta }$, where $\lambda$=wavelength and $\theta$=angle from the original beam.

Given wavelength of sodium light ${\lambda }_1=5890\mathring{A}=5890\times {10}^{-10}m$ and ${\theta }_1={0.40}^0$

Putting the above values in the given equation we get,

$d=\frac{5890\times {10}^{-10}}{\sin 0.40}=8.43\times {10}^{-5}m$

Also ${\lambda }_2=\frac{{\lambda }_1}{n}$--------(A)

where n= refractive index of water =1.33

Thus A becomes,

${\lambda }_2=\frac{5890\times {10}^{-10}}{1.33}=4.42\times {10}^{-7}m$

Again,

$d=\frac{{\lambda }_2}{\sin {\theta }_2}$

${\theta }_2={\sin }^{-1}\left(\frac{4.42\times {10}^{-7}}{1.33}\right)={0.30}^0$