In a Youngs double slit experiment, the distance between the slits is 0.1 mm and the screen is 1 m away. If the wavelength of light used is 500 nm, what is the distance between the third bright fringe and the central maximum?

B. 15 mm — The position of the nth bright fringe is given by y = n * lambda * D / d. Substituting n = 3, lambda = 500 * 10^-9 m, D = 1 m, and d = 0.1 * 10^-3 m yields 15 mm.

JEE Main · PhysicsMedium

$In a Youngs double slit experiment, the distance between the slits is $d = 0.1 \text{ mm}$ and the screen is $D = 1 \text{ m}$ away. If the wavelength of light used is $\lambda = 500 \text{ nm}$, what is the distance between the third bright fringe and the central maximum?$

A. $5 \text{ mm}$
B. $15 \text{ mm}$
C. $7.5 \text{ mm}$
D. $10 \text{ mm}$

Show correct answer & step-by-step solution

Correct answer: B — $15 \text{ mm}$

Solution

The position of the

n^{th}

bright fringe is given by

y_{n} = \frac{nλ D}{d}

. Substituting

n = 3

λ = 500 \times 1 0^{- 9} m

D = 1 m

, and

d = 0.1 \times 1 0^{- 3} m

yields

y_{3} = 15 \times 1 0^{- 3} m = 15 mm

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