For an eye kept at a depth h inside water is refractive index `mu`, and viewed outside, the radius of circle through which the outer objects can be seen will be

If we see the outside objects by keeping an eye at a depth h inside water of refractive index u, then the value of RADIUS of circle of view for outside objects will be

A bulb is kept at a depth h inside water of refractive index x . Area of bright patch on water surface is

Define critical angle. A luminous object is placed at a depth h in a medium of refractive index mu . Show that the radius r of the circular base of the one through which light can emerge is r = h // sqrt(mu^(2) - 1)

In a swimming pool, a person is viewing outside objects by keeping an eye at a depth h inside water. If the critical angle for water is ‘ theta_(c) ’, then the value of the diameter of the circle of view for outside objects will be

An air masked eye placed inside water refractive index, mu sees the outside world. The vertex angle of the cone in which the eye will see the outside world is

A point source of light is kept at a depth of 'h' in water of r.i 4/3. The radius of the circle at the surface of water through which light emits is

What is optical fibre? An object is placed at a depth ‘d’ in a medium of-refractive index mu Show that the radius of the circular base of light cone r = d/sqrt(mu^2-1)

There is a small black dot at the centre C of a solid glass sphere of refractive index mu . When seen from outside, the dot will appear to be located

There is a small black dot at the centre C of a solid glass sphere of refractive index mu . When seen from outside, the dot will appear to be located