First slide

Ray optics

Question

ABCD is a composite slab made of two slabs of transparent materials having refractive indices $μ_{1} = \frac{4}{3}$ and $μ_{2} = \frac{3}{2}$ . P is a small object. For near normal vision from top of the composite slab, the apparent position of the object P i.e. at the interface of the slabs. Then the ratio $\frac{t_{1}}{t_{2}}$ is

Moderate

A

8/3
B

7/3
C

5/8
D

8/5

Solution

Apparent depth $= \frac{t_{1}}{μ_{1}} + \frac{t_{2}}{μ_{2}}$
$∴ \frac{t_{1}}{μ_{1}} + \frac{t_{2}}{μ_{2}} = t_{1} \Rightarrow \frac{t_{1}}{t_{2}} = \frac{1 / μ_{2}}{1 - 2 / μ_{1}} = \frac{2 / 3}{1 - \frac{3}{4}} = \frac{8}{3}$

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