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Question

Physics Question on Wave optics

The two slits are 1 mm apart from each other and illuminated with a light of wavelength 5×1075 \times 10^{-7} m. If the distance of the screen is 1 m from the slits, then the distance between third dark fringe and fifth bright fringe is

A

1.5 mm

B

0.75 mm

C

1.25 mm

D

0.625 mm

Answer

1.25 mm

Explanation

Solution

Given λ=5×107\lambda = 5 \times 10^{-7} m, DD = 1 m, dd = 1 mm.
Distance of nthn^{th} bright fringe from the centre
=nDλd= \frac{n D \lambda}{d}
where n=1,2,3.....n = 1 , 2 , 3 .....
So the distance of 5th bright fringe = 5Dλd\frac{5 D \lambda}{d}
Distance of nthn^{th} dark fringe from the centre
=(n12)Dλd= \left( n - \frac{1}{2} \right) \frac{D \lambda}{d}
where n=1,2,3,4 n = 1, 2, 3, 4
3rd3^{rd} dark fringe = (312)Dλd=52Dλd\left( 3 - \frac{1}{2} \right) \frac{D \lambda}{d} = \frac{5}{2} \frac{D \lambda}{d}
=5×1×5×1072×1×103=12.5×104m=1.25mm= \frac{5 \times 1 \times 5 \times 10^{-7}}{2 \times 1 \times 10^{-3}} = 12.5 \times 10^{-4} \, m = 1.25 \, mm