Question

How does shift depend upon the thickness AD or BC of the block?

Answer 1

Hint: One of the rays is along the normal so it will go undeviated upon refraction. Another ray bends towards the normal in going inside the slab and bends away from the normal in coming out of the slab.

Complete answer:
An object O is giving out rays, one ray travels along the normal, i.e., it makes a normal incidence (makes a right angle with the slab) and another ray suffers refraction in going from rarer to denser medium and then from denser to rarer medium. The shift we are talking about is the lateral displacement l, between the first and second ray. It has to depend on the angle of incidence, the medium of refraction and also on the length which it has to travel inside the slab.

Consider the triangle PQR, it is a right angled triangle, so we may write:
$QR = AD \tan r$.
As we have already fixed the angle of incidence i, r will be fixed. Thus, the distance QR can only change if AD changes. And as this distance plus the normal to normal distance will make required lateral displacement, it is quite obvious that as AD will increase, the shift will increase.

Note:
In the right angle triangle we wrote the base to be AD, as that is the length of the slab, so it will be the same throughout. We took the shift or lateral displacement as only due to the first refraction, i.e., by slab. We did not consider the shift due to the refraction from surface CD (although ‘l’ might look like to be marked there) for measuring lateral displacement.