A small pin fixed on a table top is viewed from above from a distance of 100 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 9 cm thick glass slab held parallel to the table? Refractive index of glass =1.5.

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(3) 3 cm. Given: slab thickness t=9 cm, refractive index n=1.5. The 100 cm viewing distance is not needed. Idea: a slab with parallel faces raises whatever is seen through it by the normal shift d=t(1-1/n), and that shift does not depend on how far away the observer is. d=9(1-1/(1.5))=9×1/3. So the pin appears to be raised by 3 cm. Said another way: 9 cm of this glass looks only 6 cm thick, and the missing 3 cm is the rise.

A small pin fixed on a table top is viewed from above from a distance of 100 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 9 cm thick glass slab held parallel to the table? Refractive index of glass =1.5.

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