MCQ
For a parallel beam of monochromatic light of wavelength $'\lambda'$ diffraction is produced by a single slit whose width $'a'$ is of the order 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
  • $\frac{{2D\lambda }}{a}$
  • B
    $\;\frac{{D\lambda }}{a}$
  • C
    $\;\frac{{Da}}{\lambda }$
  • D
    $\;\frac{{2Da}}{\lambda }$

Answer

Correct option: A.
$\frac{{2D\lambda }}{a}$
a
Given situation is shown in the figure

For central maxima, $\sin \theta=\frac{\lambda}{a}$

Also, $\theta$ is very-very small so

$\sin \theta \approx \tan \theta=\frac{y}{D}$

$\therefore \quad \frac{y}{D}=\frac{\lambda}{a}, y=\frac{\lambda D}{a}$

Width of central maxima $=2 y=\frac{2 \lambda D}{a}$

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