CBSE Class 12 · 2013 · Wave OpticsShort Answer

A parallel beam of light of

450

nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen

1.5

m away. It is observed that the first minimum is at a distance of

3

mm from the centre of the screen. Calculate the width of the slit.

Answer: $a = 0.225$ mm.

Solution

For single-slit diffraction the first minimum satisfies

a sin θ = λ

, and for small angles

sin θ \approx \frac{y}{D}

, so

a \cdot \frac{y}{D} = λ \Rightarrow a = \frac{λ D}{y}

Substituting

λ = 450 \times 1 0^{- 9}

D = 1.5

y = 3 \times 1 0^{- 3}

a = \frac{450 \times 1 0 ^{- 9} \times 1.5}{3 \times 1 0 ^{- 3}} = 2.25 \times 1 0^{- 4} m

a = 0.225 mm

CBSE Class 12 Board · 2013 · Subjective