JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Wave Optics

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In a Young's double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is:

(A)4.9 mm
(B)6.9 mm
(C)3.9 mm
(D)5.9 mm✓ Correct

Answer: (D)

Solution

The spacing between successive bright fringes in a double-slit pattern is the fringe width

β = \frac{λ D}{d}

. Putting in

λ = 589 \times 1 0^{- 9}

D = 1.5

m and

d = 0.15 \times 1 0^{- 3}

m gives

β = \frac{589 \times 1 0 ^{- 9} \times 1.5}{0.15 \times 1 0 ^{- 3}} = 5.89 \times 1 0^{- 3}

m, which is close to

5.9

mm.

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option

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