$A$ mixture of light,consisting of wavelengths $590 \ nm$ and an unknown wavelength,illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further,it is observed that the $3^{rd}$ bright fringe of the known light coincides with the $4^{th}$ bright fringe of the unknown light. From this data,the wavelength of the unknown light is ...... $nm$.

  • A
    $393.4$
  • B
    $885$
  • C
    $442.5$
  • D
    $776.8$

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If a torch is used in place of monochromatic light in Young's experiment,what will happen?

Assertion: In Young's double slit experiment,the two slits are at a distance $d$ apart. An interference pattern is observed on a screen at a distance $D$ from the slits. At a point on the screen directly opposite to one of the slits,a dark fringe is observed. Then,the wavelength of the wave is proportional to the square of the distance between the two slits.
Reason: For a dark fringe,the intensity is zero.

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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 a Young's double slit experiment,the source $S$ and the two slits $A$ and $B$ are vertical,with slit $A$ above slit $B$. The fringes are observed on a vertical screen $K$. The optical path length from $S$ to $B$ is increased very slightly (by introducing a transparent material of higher refractive index) and the optical path length from $S$ to $A$ is not changed. As a result,the fringe system on $K$ moves:

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