The first minima due to a single slit diffraction is at $\theta = 30^o$ for a light of wavelength $5000 \, \mathring{A}$. The width of the slit is

  • A
    $5 \times 10^{-5} \, cm$
  • B
    $1.0 \times 10^{-4} \, cm$
  • C
    $2.5 \times 10^{-5} \, cm$
  • D
    $1.25 \times 10^{-5} \, cm$

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Similar Questions

Which of the following statements are correct with reference to single slit diffraction pattern?
$(I)$ The central maxima is twice as wide as the secondary maxima.
$(II)$ The intensity of secondary maxima decreases as we move away from the central maxima.
$(III)$ The width of the central maxima is independent of the slit width.
$(IV)$ The intensity of the central maxima is the same as that of the secondary maxima.

If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is $30^{\circ}$ in a single slit diffraction pattern recorded using $628 \ nm$ light,then the width of the slit is . . . . . . $\mu m$.

$A$ single slit diffraction pattern is formed with white light. For what wavelength of light the $4^{\text{th}}$ secondary maximum in the diffraction pattern coincides with the $3^{\text{rd}}$ secondary maximum in the pattern of light of wavelength $\lambda$?

The Fraunhofer diffraction pattern of a single slit is formed in the focal plane of a lens of focal length $1\,m$. The width of the slit is $0.3\,mm$. If the third minimum is formed at a distance of $5\,mm$ from the central maximum, then the wavelength of light will be......$\mathring{A}$

For the visibility of a diffraction pattern of a wave of light,the size of the slit or obstacle should be

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