SATHEE: Wave Optics - Result Question 28

Wave Optics - Result Question 28

30. For a parallel beam of monochromatic light of wavelength ’ $λ$ ‘, 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) $\frac{D λ}{a}$

(b) $\frac{D a}{λ}$

(c) $\frac{2 D a}{λ}$

(d) $\frac{2 D λ}{a}$

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Solution:

(d) Linear width of central maxima $Y$

$= D (2 θ) = 2 D θ = \frac{2 D λ}{a} ∴ θ = \frac{λ}{a}$

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

The angular width ’ $d$ ’ central maxima $= 2 θ = \frac{2 λ}{b}$

Linear width of central maxima $= 2 D θ = \frac{2 D λ}{b}$

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

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Wave Optics - Result Question 28

30. For a parallel beam of monochromatic light of wavelength ’ λ ‘, 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]

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30. For a parallel beam of monochromatic light of wavelength ’ $λ$ ‘, 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]