$A$ small bulb is placed at the bottom of a tank containing water to a depth of $80 \; cm$. What is the area (in $m^2$) of the surface of water through which light from the bulb can emerge out? Refractive index of water is $1.33$. (Consider the bulb to be a point source.)

The critical angle for light traveling from medium $(i)$ to medium $(ii)$ is $\theta$. If the speed of light in medium $(i)$ is $v$,then the speed of light in medium $(ii)$ is:

$A$ normally incident ray is reflected at an angle of $90^o$. The value of the critical angle is.......$^o$

Assertion : In optical fibre,the diameter of the core is kept small.
Reason : This smaller diameter of the core ensures that the fibre should have incident angle more than the critical angle required for total internal reflection.

$A$ planar structure of length $L$ and width $W$ is made of two different optical media of refractive indices $n_1=1.5$ and $n_2=1.44$ as shown in the figure. If $L \gg W$,a ray entering from end $AB$ will emerge from end $CD$ only if the total internal reflection condition is met inside the structure. For $L = 9.6 \ m$,if the incident angle $\theta$ is varied,the maximum time taken by a ray to exit the plane $CD$ is $t \times 10^{-9} \ s$,where $t$ is. . . . . . . [Speed of light $c = 3 \times 10^8 \ m/s$]

With respect to air,the critical angle in a medium for light of red colour $[{\lambda _1}]$ is $\theta$. Other factors remaining the same,the critical angle for light of yellow colour $[{\lambda _2}]$ will be: