$A$ beam of monoenergetic electrons,which have been accelerated from rest by a potential $U$,is used to form an interference pattern in a Young's double slit experiment. The electrons are now accelerated by potential $4U$. Then,the fringe width

  • A
    remains same
  • B
    is half the original fringe width
  • C
    is twice the original fringe width
  • D
    is one-fourth the original fringe width

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Similar Questions

Two light waves of wavelengths $800 \, nm$ and $600 \, nm$ are used in Young's double slit experiment to obtain interference fringes on a screen placed $7 \, m$ away from the plane of the slits. If the two slits are separated by $0.35 \, mm$,then the shortest distance from the central bright maximum to the point where the bright fringes of the two wavelengths coincide will be $............. \, mm$.

In a Young's double slit experiment,a student observes $8$ fringes in a certain segment of the screen when a monochromatic light of $600 \ nm$ wavelength is used. If the wavelength of light is changed to $400 \ nm$,then the number of fringes he would observe in the same region of the screen is:

In Young's double slit experiment, the distance between the two slits is $0.1 \, mm$ and the wavelength of light used is $4 \times 10^{-7} \, m$. If the width of the fringe on the screen is $4 \, mm$, the distance between the screen and the slit is:

In Young's double-slit experiment,the slits are separated by $0.2 \, cm$ and illuminated by light of wavelength $\lambda = 5896 \, \mathring{A}$. What is the fringe width on a screen placed $1 \, m$ away from the plane of the slits? What will be the fringe width if the entire system is immersed in water (in $, mm$)? (Refractive index of water $\mu_w = 4/3$)

In Young's double slit experiment,the wavelength of light was changed from $7000 \ \mathring{A}$ to $3500 \ \mathring{A}$. While doubling the separation between the slits,which of the following is not true for this experiment?

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