The figure shows a double-slit experiment. Each slit has a width $W$. $A$ thin glass slab of thickness $\delta$ and refractive index $\mu$ is placed in front of one of the slits. The intensity at the central point $C$ is measured as a function of the thickness $\delta$. At what thickness $\delta$ will the intensity at $C$ be minimum?

  • A
    $\frac{(2n - 1)\lambda}{2}$
  • B
    $\frac{\lambda}{2(\mu - 1)}$
  • C
    $\frac{(2n - 1)\lambda}{2(\mu - 1)}$
  • D
    $\frac{2\lambda}{(\mu - 1)}$

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In a double-slit experiment,the wavelength $\lambda$ of the light source is $400 \ nm$,the slit separation $d$ is $20 \ \mu m$,and the individual slit width $a$ is $4 \ \mu m$. Considering the interference of light from the two slits and the diffraction of light through each slit,find the number of bright interference fringes that lie within the central peak of the diffraction envelope.

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In the figure shown,if a parallel beam of light is incident on the plane of the slits at an angle such that the path difference at point $O$ is $\Delta x = d \sin \theta = d \cdot (\frac{2d/3}{d}) = \frac{2d}{3}$,and if point $O$ is a maxima for monochromatic light,then which of the following cannot be the wavelength of the incident light? [Assume $d << D, \lambda << d$]

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:

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