A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? (Refractive index of glass =1.5)

Question

Accepted Answer

(1) 5 cm. Given: slab thickness t=15 cm, n=1.5; the 50 cm viewing distance plays no part. Idea: a parallel slab raises what is under it by the normal shift d=t(1-1/n), which for n=1.5 is one third of the thickness. d=15(1-1/(1.5))=15×1/3. So the pin appears raised by 5 cm. Equivalently, the 15 cm of glass looks only 10 cm thick.

A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? (Refractive index of glass =1.5)

Step-by-step solution

Why the other options are wrong