How to solve this question?

Solution : 

Actual depth of the bulb in water, D = h cm 

Refractive index of water, μ = 4/3 => 1.33

The given situation is shown in the following figure:

Where,

i = Angle of incidence

r = Angle of refraction = 90°

Since the bulb is a point source, the emergent light can be considered as a circle of radius,
R = AC/2 = AO = OB

Using Snell’ law, we can write the relation for the refractive index of water as:
μ = sin r/sin i 
1.33 = sin 90°/sin i 
i = sin^-1 (1/1.33) 
i = 48.75° 

Using the given figure, we have the relation:
tan i = OC/OB = R/D 
tan 48.75° = R/h 

∴ R = tan 48.75° × h 
R = 1.14 h or 3 h/√7                 (option a)