Question

In Young's double slit experiment the two slits are illuminated by light of wavelength 5890${}^{\circ }$A and the distance between the fringes obtained on the screen is 0.2${}^{\circ }$. If the whole apparatus is immersed in water then the angular fringe width will be (refractive index of water is 4/3):

• A. ${0.30}^{\circ }$
• B. ${0.15}^{\circ }$
• C. ${15}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

Answer: (B)

${0.15}^{\circ }$

For interference in YDSE
$dsin\theta =n\lambda$
for small $\theta ,sin\theta$ can be approximated to $\theta$
$So,d\theta =n\lambda$
$So,\theta \alpha \lambda$
Now as the set up immersed in water
$\lambda$ will become $\frac{\lambda }{\mu }=\frac{\lambda }{4/3}=\frac{3\lambda }{4}$
$So,\frac{{\lambda }_1}{{\lambda }_2}=\frac{{\theta }_1}{{\theta }_2}$
$\frac{\lambda }{3\lambda /4}=\frac{{0.2}^{\circ }}{{\theta }_2}$
${\theta }_2={0.2}^{\circ }\times \frac{3}{4}={0.15}^{\circ }.$