In Young's double slit experiment, the slits are 0.5 , mm apart a… - Vidyadip

MCQ

In Young's double slit experiment, the slits are $0.5\, mm$ apart and interference pattern is observed on a screen placed at a distance of $1.0\, m$ from the plane containing the slits. If wavelength of the incident light is $6000 Å$, then the separation between the third bright fringe and the central maxima is......$mm$

A
$4$
✓
$3.5$
C
$3$
D
$2.5$

✓

Answer

Correct option: B.

$3.5$

b
(b)Separation ${n^{th}}$bright fringe and central maxima is ${x_n} = \frac{{n\lambda D}}{d}$
So, ${x_3} = \frac{{3 \times 6000 \times {{10}^{ - 10}} \times 1}}{{0.5 \times {{10}^{ - 3}}}} = 3.5\,mm.$

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