Doubt & Answer Explanation

Answer Explanation

Lets consider the two cases. In the first case the ray is going from the denser to the rarer medium while in the second case it is going from the rarer to the denser medium.




Here we have two cases, in the first figure ray goes from denser to rarer medium i.e., u₂ < u₁ while in the second figure u₂ > u₁.

In both cases , we can apply snail's law as:

u₁sini = u₂sinr

If angles are small we get:

u₁tani = u₂tanr .  ............... (i)

In both cases, the real depth is BO while the apparent depth is BI.


Thus, using (i) we get:

u₂/ u₁ = tani/tanr = (BA/OB)  / (BA / BI)  = BI / BO  = (apparent depth) / (real depth).


Now, in your case we are watching from a denser medium, so figure (ii) will represent your question.

So, if  in figure (ii) u₁ is R.I. of air then we can put u₁ = 1 and u₂ = u.
So, we get :

u = (apparent depth) / (real depth).

It can be verified too , as we know that u is more than one (refractive index of a medium denser than air).
And the apparent depth is more than real depth too as when we produce the refracted ray backwards it meets the normal at a point behind the point the light actually imerges from.

I would suggest that , you should always draw the ray diagram in these cases rather than applying the direct formula.

Real depth is the depth of the object while apparent depth is the depth of the image.