Question

In a double slit experiment the angular width of a fringe is found to be 0.2⁰on a screen placed I m away. The wavelength of light used is 600nm. The angular width of the fringe if entire experimental apparatus is immersed in water is $\left( Take\mu_{watre} = \frac{4}{3} \right)$

• A. 0.15⁰
• B. 1⁰
• C. 2⁰
• D. 0.3⁰

Answer 1

Answer: (A)

0.15⁰

Answer 2

Since a fringe of width $\beta$is formed on the screen at distance D from the slits, so the angular fringe width

$\theta = \frac{\beta}{D} = \frac{D\lambda/d}{D} = \frac{\lambda}{d}$ $\lbrack\because\beta = \frac{D\lambda}{d}\rbrack$

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

If the wavelength in water be $\lambda'$ and the angular fringe width be $\theta'$ then

$d = \frac{\lambda'}{\theta'}or\frac{\lambda}{\theta} = \frac{\lambda'}{\theta'}$

Or $\theta' = \frac{\lambda'}{\lambda}.\theta = \frac{\lambda/\mu}{\lambda}.\theta$ $\lbrack\because\lambda'\frac{\lambda}{\mu}\rbrack$

$= \frac{0.2º}{4/3} = 0.15º$