The diffraction of light starts from an obstacle (edge). It depends on ...

  • A
    Size of the obstacle (edge)
  • B
    Size of the obstacle (edge) and wavelength
  • C
    Distance of the screen from the obstacle (edge) and wavelength
  • D
    Size of the obstacle and its distance from the screen

Explore More

Similar Questions

Light of wavelength $6000 \, \mathring A$ is incident on a slit of width $0.30 \, mm$. $A$ screen is placed at a distance of $2 \, m$ from the slit. Find the position of the first minimum.

Difficult
View Solution

Light of wavelength $\lambda$ is incident on a single slit of width $a$,and the distance between the slit and the screen is $D$. In the diffraction pattern,if the slit width is equal to the width of the central maximum,then $D=$

Light of wavelength $550\, nm$ falls normally on a slit of width $22.0 \times 10^{-5}\, cm$. The angular position of the second minima from the central maximum will be (in radians)

In the diffraction pattern of a single slit,the width of the secondary maxima compared to the central maximum is:

In the case of Fraunhofer diffraction at a single slit,which of the following statements correctly describes the diffraction pattern on the screen?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo