$A$ double slit experiment is performed with light of wavelength $500\, nm$. $A$ thin film of thickness $2\, \mu m$ and refractive index $1.5$ is introduced in the path of the upper beam. The location of the central maximum will

  • A
    Remain unshifted
  • B
    Shift downward by nearly two fringes
  • C
    Shift upward by nearly two fringes
  • D
    Shift downward by $10$ fringes

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Similar Questions

The central fringe of the interference pattern produced by light of wavelength $6000 \mathring A$ is found to shift to the position of the $4^{\text {th}}$ bright fringe after a glass plate of refractive index $1.5$ is introduced in front of one slit in Young's experiment. The thickness of the glass plate will be ......... $\mu m$.

The figure shows a Young's double slit experimental setup. It is observed that when a thin transparent sheet of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits,the central maximum shifts by a distance equal to $n$ fringe widths. If the wavelength of light used is $\lambda$,then $t$ will be:

$6300 \mathring A$ wavelength light shines on two narrow slits separated by a distance of $1.0 \ mm$ and illuminates a screen at a distance of $1.5 \ m$ away. When one slit is covered by a thin glass plate of refractive index $1.8$ and the other slit by a thin glass plate of refractive index $\mu$,the central maxima shifts by $6^o$. Both plates have the same thickness of $0.5 \ mm$. The value of the refractive index $\mu$ of the plate is:

On replacing a thin film of mica of thickness $12 \times 10^{-5} \ cm$ in the path of one of the interfering beams in Young's double slit experiment using monochromatic light,the fringe pattern shifts through a distance equal to the width of bright fringe. If $\lambda = 6 \times 10^{-5} \ cm$,the refractive index of mica is:

$A$ monochromatic light source $S$ of wavelength $440 \,nm$ is placed slightly above a plane mirror $M$ as shown below. The image of $S$ in $M$ can be used as a virtual source to produce interference fringes on the screen. The distance of source $S$ from $O$ is $20.0 \,cm$ and the distance of the screen from $O$ is $100.0 \,cm$ (figure is not to scale). If the angle $\theta = 0.50 \times 10^{-3} \,radians$, then the width of the interference fringes observed on the screen is ............... $mm$.

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