Question

In Young's double slit experiment, the two slits are illuminated by light of wavelength $5890\stackrel{\circ }{A}$ and the angular fringe width is ${0.2}^{\circ }$. If the whole apparatus is immersed in water, then the new angular fringe width will be (refractive index of water $=\frac{4}{3}$).

• A. ${0.05}^{\circ }$
• B. ${0.20}^{\circ }$
• C. ${0.35}^{\circ }$
• D. ${0.15}^{\circ }$

Answer 1

The correct option is D ${0.15}^{\circ }$


Angular fringe width $\varphi =\frac{\beta }{D}$ (for small angles)
$=\frac{\lambda D}{d}\times \frac{1}{D}=\frac{\lambda }{d}$

Thus, $\frac{{\varphi }_w}{{\varphi }_a}=\frac{{\lambda }_w}{{\lambda }_a}$

Wavelength of light in water ${\lambda }_w=\frac{{\lambda }_a}{\mu }$
$\therefore {\varphi }_w=\frac{{\varphi }_a}{\mu }=\frac{0.2}{\frac{4}{3}}={0.15}^{\circ }$