Wave OpticsHard
Question
In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness $t$ and refractive index $n( = 1.5)$, the central fringe shifts by 0.2 cm . The value of t is $\_\_\_\_$ cm .
Options
A.$8 \times 10^{- 4}$
B.$6.0 \times 10^{- 3}$
C.$5.6 \times 10^{- 4}$
D.$5.0 \times 10^{- 3}$
Solution
$dsin\theta = (\mu - 1)t$
$$d\left\lbrack \frac{x}{D} \right\rbrack = (\mu - 1)t$$
$$\begin{matrix} t & \ = \frac{xd}{D(\mu - 1)} \\ & \ = \frac{(0.2)(0.1)}{50(1.5 - 1)} \end{matrix}$$
$$t = 8 \times 10^{- 4}\text{ }cm$$
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