Wave Optics - Result Question 28

30. For a parallel beam of monochromatic light of wavelength ’ $\lambda$ ‘, diffraction is produced by a single slit whose width ’ $a$ ’ is of the wavelength of the light. If ’ $D$ ’ is the distance of the screen from the slit, the width of the central maxima will be :[2015]

(a) $\frac{D \lambda}{a}$

(b) $\frac{Da}{\lambda}$

(c) $\frac{2 Da}{\lambda}$

(d) $\frac{2 D \lambda}{a}$

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Solution:

  1. (d) Linear width of central maxima $Y$

$=D(2 \theta)=2 D \theta=\frac{2 D \lambda}{a} \quad \therefore \theta=\frac{\lambda}{a}$

The central maxima lies between the first minima on both sides.

The angular width ’ $d$ ’ central maxima $=2 \theta=\frac{2 \lambda}{b}$

Linear width of central maxima $=2 D \theta=\frac{2 D \lambda}{b}$

$b=$ width of slits and $D=$ distance between slit and screen.



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