In a Young's double slit experiment, the distance between the slits is 0.1 mm and the distance between the screen and the slits is 1 m. If the light source has a wavelength of 500 nm, what is the distance between the third bright fringe and the central maximum?

C. 15 mm — The position of the third bright fringe is calculated using the formula for fringe width multiplied by the order.

JEE Main · PhysicsMedium

$In a Young's double slit experiment, the distance between the slits is d = 0.1 mm and the distance between the screen and the slits is D = 1 m . If the light source has a wavelength of λ = 500 nm, what is the distance y_{3} between the third bright fringe and the central maximum?$

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Correct answer: C — $15 mm$

$The position of the n -th bright fringe is given by y_{n} = \frac{nλ D}{d} .$
$Substitute the given values: n = 3, λ = 500 \times 1 0^{- 9} m, D = 1 m, and d = 0.1 \times 1 0^{- 3} m .$
$y_{3} = \frac{3 \times 500 \times 1 0 ^{- 9} \times 1}{0.1 \times 1 0 ^{- 3}} = 15 \times 1 0^{- 3} m .$
$Hence the answer is (C).$