JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Wave Optics

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In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of

5890 \overset{˚}{A}

is:

(A) $1178 \times 1 0^{- 6} m$ ✓ Correct
(B) $1178 \times 1 0^{- 12} m$
(C) $1178 \times 1 0^{- 9} m$
(D) $5890 \times 1 0^{- 7} m$

Answer: (A)

Solution

The fringe width in Young's arrangement is

β = \frac{λ D}{d} = \frac{5890 \times 1 0 ^{- 10} \times 0.5}{0.5 \times 1 0 ^{- 3}} = 5.89 \times 1 0^{- 4} m

Successive bright fringes are separated by

β

, so the first and the third bright fringes are two fringe widths apart:

Δ x = 2 β = 2 \times 5.89 \times 1 0^{- 4} = 1.178 \times 1 0^{- 3} m

The distance between the first and the third bright fringe is

1178 \times 1 0^{- 6}

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Wave Optics

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