First slide

Diffraction

Question

A linear aperture whose width is $0.02 c m$ is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength $5 \times 10^{- 5} c m$ . The distance of the first dark band of the diffraction pattern from the centre of the screen is

Moderate

A

0.10 cm
B

0.25 cm
C

0.20 cm
D

0.15 cm

Solution

Here,
$\begin{array}{l} a = 0.02 cm = 2 \times 10^{- 4} m \\ λ = 5 \times 10^{- 5} cm = 5 \times 10^{- 7} m \\ D = 60 cm = 0.6 m \end{array}$
Position of first minima on the diffraction pattern,
$y_{1} = \frac{D λ}{a} = \frac{0.6 \times 5 \times 10^{- 7}}{2 \times 10^{- 4}} = 15 \times 10^{- 4} m = 0.15 cm$

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