In $YDSE$,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. Then,the fringe width:

  • A
    becomes $\frac{1}{20}$ of the original value
  • B
    becomes $\frac{1}{90}$ of the original value
  • C
    remains the same
  • D
    becomes $\frac{1}{10}$ of the original value

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Similar Questions

$A$ mixture of light of wavelength $590 \, nm$ and an unknown wavelength is incident on the two slits of a Young's double-slit experiment. The central bright fringes of both lights coincide. The $3^{rd}$ bright fringe of the known wavelength coincides with the $4^{th}$ bright fringe of the unknown wavelength. Find the unknown wavelength in $nm$.

In the ideal double-slit experiment,when a glass plate (refractive index $\mu = 1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$),the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is:

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In Young's double slit experiment,how many maximas can be obtained on a screen (including the central maximum) on both sides of the central fringe if $\lambda = 2000\,{\mathring A}$ and $d = 7000\,{\mathring A}$?

Young's experiment demonstrates that . . . . . .

In Young's double slit experiment,if one of the slits is closed fully,then in the interference pattern:

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