Question

In a double slit experiment the angular width of a fringe is found to be 0.2 on a screen placed 1 m away. The wavelength of light used is 600 nm. The angular width of the fringe if entire experimental apparatus is immersed in water is $(Take{\mu }_{water}=\frac{4}{3})$

• A. ${0.15}^{\circ }$
• B. $1^{\circ }$
• C. $2^{\circ }$
• D. ${0.3}^{\circ }$

Answer 1

Answer: (A)

${0.15}^{\circ }$

Since a fringe of width $\beta$ is formed on the screen at distance D from the slits, so the angular fringe width
$\theta =\frac{\beta }{D}=\frac{D\lambda /d}{D}=\frac{\lambda }{d}$ $[\therefore \beta =\frac{D\lambda }{d}]$
$\Rightarrow d=\frac{\lambda }{\theta }$
If the wavelength in water be $\lambda$ and the angular fringe width be ${\theta }^{\prime },$ then
$d=\frac{{\lambda }^{\prime }}{{\theta }^{\prime }}or\frac{\lambda }{\theta }=\frac{{\lambda }^{\prime }}{{\theta }^{\prime }}or{\theta }^{\prime }=\frac{{\lambda }^{\prime }}{\lambda }\cdot \theta =\frac{\lambda /\mu }{\lambda }\cdot \theta [\therefore {\lambda }^{\prime }=\frac{\lambda }{\mu }]$

$\therefore {\theta }^{\prime }=\frac{{0.2}^{\circ }}{4/3}={0.15}^{\circ }$