$A$ light ray falls on a square glass slab as shown in the diagram. The index of refraction of the glass,if total internal reflection is to occur at the vertical face,is equal to

  • A
    $\frac{(\sqrt{2} + 1)}{2}$
  • B
    $\sqrt{\frac{5}{2}}$
  • C
    $\frac{3}{2}$
  • D
    $\sqrt{\frac{3}{2}}$

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$A$ cylindrical transparent rod of refractive index $\mu = \frac{2}{\sqrt{3}}$ is surrounded by air. $A$ light ray is incident at the midpoint of one end of the rod as shown in the figure. The angle of incidence $\theta$ for which the light ray undergoes total internal reflection at the side wall of the rod is:

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Four rays $1, 2, 3$ and $4$ are incident normally on the face $PQ$ of an isosceles prism $PQR$ with apex angle $\angle Q = 120^{\circ}$. The refractive indices of the material of the prism for the above rays $1, 2, 3$ and $4$ are $1.85, 1.95, 2.05$ and $2.15$ respectively and the surrounding medium is air. Then,the rays emerging from the face $QR$ are

Glass has refractive index $\mu$ with respect to air and the critical angle for a ray of light going from glass to air is $\theta$. If a ray of light is incident from air on the glass with angle of incidence $\theta$,the corresponding angle of refraction is:

$A$ diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is $\frac{4}{3}$,and the diver's eyes are $15 \, cm$ below the surface of water. Then the radius of the circle is

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