CBSE Class 12 · Wave OpticsLong Answer

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(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence obtain the conditions for the angular width of the secondary maxima and secondary minima. (b) Two wavelengths of sodium light of

590

nm and

596

nm are used in turn to study the diffraction taking place at a single slit of aperture

2 \times 1 0^{- 6}

m. The distance between the slit and the screen is

1.5

m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases.

Solution

(a) A plane wavefront on a single slit

A B

(width

a

) sends out secondary wavelets (Huygens). Those along

θ = 0

reinforce to give the central maximum. For a direction

θ

, the edge-to-edge path difference is

a sin θ

, giving:

Secondary minima: a sin θ = nλ, n = 1, 2, 3, \dots

Secondary maxima: a sin θ = (2 n + 1) \frac{λ}{2}, n = 1, 2, 3, \dots

(b) The first secondary maximum is at

y_{1} = \frac{3 λ D}{2 a}

. The separation for the two wavelengths is

Δ y = \frac{3 D}{2 a} (λ_{2} - λ_{1}) = \frac{3 \times 1.5}{2 \times 2 \times 1 0 ^{- 6}} (596 - 590) \times 1 0^{- 9}

Δ y = \frac{4.5}{4 \times 1 0 ^{- 6}} \times 6 \times 1 0^{- 9} = 6.75 \times 1 0^{- 3} m

Δ y = 6.75 mm

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CBSE Class 12 · Wave OpticsLong Answer

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