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In a double slit experiment , when light of wavelength 400 nm was used , the angular width of the first minima formed on a screen palced 1 m away , was found to be $0.2 \degree$ . what will be the angular width of the first minima , if the entire experimental apparatus is immersed in water ?

$0.15 \degree$

$0.05 \degree$

$0.1 \degree$

Answers (1)

best_answer

Refractive index -

$mu = frac{c}{v}$

- wherein

$c=$ speed of light in vacuum.

$v=$ speed of light in medium.

Young Double Slit Experiment -

- wherein

$y=Delta xcdot left ( frac{D}{d} ight )$

$y=$ Distance of a point on screen from central maxima

$Delta x=$ Path difference at that point

Fringe Width in a medium -

${\beta }'= \frac{\beta }{\mu }$

- wherein

Since wavelength in a medium become $\lambda = \frac{\lambda _{o}}{\mu }$

Angular width = $\theta \ \alpha \ \ \lambda$

and $\mu =\frac{c}{v}=\frac{\lambda _1}{\lambda _2}=\frac{\theta _1}{\theta _2}$

$\theta _2=\frac{\theta _1 }{\mu }\Rightarrow \frac{0.2*3}{4}=0.15^0$

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