In a single slit diffraction experiment, the width of the slit is 0.2 mm and the screen is 2 m away. If the wavelength of light is 600 nm, what is the width of the central maximum?

B. 12 mm — The width of the central maximum in a single slit diffraction pattern is given by 2 * lambda * D / a. Substituting the given values results in 12 mm.

JEE Main · PhysicsMedium

$In a single slit diffraction experiment, the width of the slit is $a = 0.2 \text{ mm}$ and the screen is $D = 2 \text{ m}$ away. If the wavelength of light is $\lambda = 600 \text{ nm}$, what is the width of the central maximum?$

A. $6 \text{ mm}$
B. $12 \text{ mm}$
C. $3 \text{ mm}$
D. $24 \text{ mm}$

Show correct answer & step-by-step solution

Correct answer: B — $12 \text{ mm}$

Solution

The width of the central maximum is

w = \frac{2 λ D}{a}

. Substituting

λ = 600 \times 1 0^{- 9} m

D = 2 m

, and

a = 0.2 \times 1 0^{- 3} m

gives

w = \frac{2 \times 600 \times 1 0 ^{- 9} \times 2}{0.2 \times 1 0 ^{- 3}} = 12 \times 1 0^{- 3} m = 12 mm

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