First slide

Young's double slit Experiment

Question

In Young's double slit experiment when wavelength used is $6000 Å$ and the screen is 40 cm from the slits, the fringes are 0.012 cm wide. What is the distance between the slits?

Easy

A

0.024 cm
B

2.4 cm
C

0.24 cm
D

0.2 cm

Solution

$β = \frac{λD}{d}$
$\Rightarrow d = \frac{λD}{β}$
$= \frac{6000 \times 10^{- 10} \times (40 \times 10^{- 2})}{0 .012 \times 10^{- 2}} = 0 .2 cm .$

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