JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Wave Optics

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Sodium light of wavelengths 650 nm and 655 nm is used to study diffraction at a single slit of aperture 0.5 mm. The distance between the slit and the screen is 2.0 m. The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is ______

\times 1 0^{- 5}

Answer: 3

Solution

For the first maximum in single-slit diffraction, the condition is

a sin θ = \frac{3 λ}{2}

. For small angles,

sin θ \approx θ = \frac{y}{L}

, so

y = \frac{3 λ L}{2 a}

.

For the two wavelengths:

$y_{1} = \frac{3 λ _{1} L}{2 a}$
$y_{2} = \frac{3 λ _{2} L}{2 a}$

The separation is:

Δ y = y_{2} - y_{1} = \frac{3 L ( λ _{2} - λ _{1} )}{2 a}

Substituting values:

λ_{1} = 650

nm,

λ_{2} = 655

nm,

L = 2

a = 0.5

mm:

Δ y = \frac{3 \times 2 \times ( 655 - 650 ) \times 1 0 ^{- 9}}{2 \times 0.5 \times 1 0 ^{- 3}} = \frac{3 \times 2 \times 5 \times 1 0 ^{- 9}}{1 0 ^{- 3}} = 30 \times 1 0^{- 6} = 3 \times 1 0^{- 5} m

Therefore, the answer is 3.

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Wave Optics

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