Question

A pin placed in contact with one surface of a glass block of thickness 15ck and of refractive index 1.5 is viewed by the normal of the opposite surface . By how much does the pin appear to be displaced .

Answer 1

The refractive index can be found with the following equation:

$$n=\frac{d_\text{real}}{d_\text{apparent}},\\\space\\ d_\text{apparent}=\frac{d_\text{real}}{n}=\frac{15}{1.5}=10\text{ cm}.$$

So, the pin appear to be displaced by

$$d=d_\text{real}-d_\text{apparent}=15-10=5\text{ cm}.$$