Pin is closer!

Classical Mechanics Level 3

A small pin on a flat table is viewed from 30 cm 30\text{ cm} directly above. If the pin is viewed from the same point through a glass slab of thickness 10 cm 10\text{ cm} and refractive index 1.6, how far it appears to be raised?


The answer is 3.75.

Samara Simha Reddy by Samara Simha Reddy , 22, India

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1 solution

μ = real thickness apparent thickness raised distance = real thickness − apparent thickness = real thickness − real thickness μ = 10 − 10 1.6 = 3.75 c m . \large \displaystyle \mu = \frac{\text{real thickness}}{\text{apparent thickness}}\\ \begin{aligned} \large \text{raised distance} = \text{real thickness} - \text{apparent thickness}\\ \large = \text{real thickness} - \frac{\text{real thickness}}{\mu}\\ \large \displaystyle = 10 - \frac{10}{1.6} \large \displaystyle = \boxed{3.75} cm.\end{aligned}

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