In a double-slit experiment, when light of wavelength $400 ext{ nm}$ was used, the angular width of the firs

Question

In a double-slit experiment, when light of wavelength $400 	ext{ nm}$ was used, the angular width of the first minima formed on a screen placed $1 	ext{ m}$ away, was found to be $0.2^{\circ}$. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? $\left(\mu_{	ext{water}} = \frac{4}{3}ight)$

Accepted Answer

The angular width of fringes in Young's Double Slit Experiment is given by $	heta = \frac{\lambda}{d}$.
When the apparatus is immersed in a liquid of refractive index $\mu$, the wavelength of light in the liquid becomes $\lambda' = \frac{\lambda}{\mu}$.
Therefore, the new angular width $	heta'$ will be:
$	heta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} = \frac{	heta}{\mu}$
Given, initial angular width $	heta = 0.2^{\circ}$ and refractive index of water $\mu = \frac{4}{3}$.
Substituting the values:
$	heta' = \frac{0.2^{\circ}}{4/3} = 0.2^{\circ} 	imes \frac{3}{4} = 0.15^{\circ}$.
Thus, the new angular width of the first minima will be $0.15^{\circ}$.

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