A pond of clear water appears to be shallower than it really is. Similarly, the pebbles lying at the bottom of the clear water of a river appear to be raised up above their actual position.
Point to Ponder
why does the apparent depth of the pond look less than its real depth?
Relation of refractive index to real and apparent depth
In fig. 14.7, o is a bright point below a glass slab. A ray OA from O perpendicular to the surface of the medium passes straight into air. We consider another ray OB from O which is refracted along BC at the point ‘b’ because air is a rarer medium so angle of refraction is greater than the angle of incidence. When the ray CBis produced backwards, it meets the first ray OA at the point I. for an observer’s eye the object O appears as it were at I above O. thus I is the virtual image of object O.
Suppose refractive index of glass is ‘n’ with respect to air. It will have this value in case when light is entering into the glass from air, but in the present case, the light is entering from glass into the air. In this case, we can find the refractive index of air with respect to glass by using snell’s law, its value with be 1/n. Hence
When B is very close to A i.e., observer’s eye is looking both the rays O A and OB, then IB = IA and OB = OA
Example 14.3: the real depth of a swimming pool is 2m. What is the apparent depth of the pool if the refractive index of water is 1.33?
Solution:
Real depth =2m n= 1.33 , apparent depth =?
= 1. 5 m