In a Young’s experiment, one of the slits is covered with a transparen’t sheet of thickness $3.6 imes 10^{- 3}$ cm due to which position of central fringe shifts to a position originally occupied by 30$^{th}$ fringe. The refractive index of the sheet, if $\lambda = 6000\ \mathring{\mathrm{A}}$, is

Question

In a Young’s experiment, one of the slits is covered with a transparen’t sheet of thickness $3.6 	imes 10^{- 3}$ cm due to which position of central fringe shifts to a position originally occupied by 30$^{th}$ fringe. The refractive index of the sheet, if $\lambda = 6000\ \mathring{\mathrm{A}}$, is

Dr. Abubacker · Accepted Answer

(a) The position of 30$^{th}$ bright fringe
$y_{30} = \frac{30\lambda D}{d}$
Now position shift of central fringe is
$y_{0} = \frac{30\lambda D}{d}$
But we know, $y_{0} = \frac{D}{d}(\mu - 1)t$
$\frac{30\lambda D}{d} = \frac{D}{d}(\mu - 1)t$
$(\mu - 1) = \frac{30\lambda}{t} = \frac{30 	imes \left( 6000 	imes 10^{- 10} ight)}{\left( 3.6 	imes 10^{- 5} ight)} = 0.5$
$	herefore\ \ \mu = 1.5$

Explanation

Reviewed by Dr. Abubacker

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