The minimum area of the disc needed to cut off all the light coming out of water $\left(\mu_{w} = \frac{4}{3}\right)$ from a point source of light placed $7 \ m$ below the water surface is: (in $\pi \ m^2$)

$A$ step index fibre has a relative refractive index difference of $0.88\%$. What is the critical angle at the core-cladding interface in degrees?

If $\mu_1$ and $\mu_2$ are the refractive indices of the materials of core and cladding of an optical fibre,then the loss of light due to its leakage can be minimised by having

In the figure,$ABC$ is the cross-section of a right-angled prism and $BCDE$ is the cross-section of a glass slab. The value of $\theta$ so that light incident normally on the face $AB$ does not cross the face $BC$ is (given $\sin^{-1}(3/5) = 37^o$):

For total internal reflection to take place,the angle of incidence $i$ and the refractive index $\mu$ of the medium must satisfy the inequality

The figure shows a ray incident at an angle $i = \pi /3.$ If the plot drawn shows the variation of $|r - i|$ versus $\frac{\mu_1}{\mu_2} = k,$ ($r =$ angle of refraction) choose the < strong>wrong alternative.