A rectangular slab of refractive index $μ$ is placed over another slab of refractive index 3, both slabs being identical in dimensions. If a coin is placed below the lower slab, for what value of $μ$ will the coin appear to be placed at the interface between the slabs when viewed from the top

1.8

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1.5

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2.5

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Solution

The correct option is C 1.5
Apparent depth of coin as seen from top = $\frac{x}{μ_{1}} + \frac{x}{μ_{2}} = x$
$\Rightarrow \frac{x}{μ_{1}} + \frac{x}{μ_{2}} = 1 \Rightarrow \frac{1}{3} + \frac{1}{μ} = 1 \Rightarrow μ = 1.5$

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A rectangular slab of refractive index μ is placed over another slab of refractive index 3, both slabs being identical in dimensions. If a coin is placed below the lower slab, for what value of μ will the coin appear to be placed at the interface between the slabs when viewed from the top

The correct option is C 1.5Apparent depth of coin as seen from top =xμ1+xμ2=x ⇒xμ1+xμ2=1⇒13+1μ=1⇒μ=1.5

A rectangular slab of refractive index $μ$ is placed over another slab of refractive index 3, both slabs being identical in dimensions. If a coin is placed below the lower slab, for what value of $μ$ will the coin appear to be placed at the interface between the slabs when viewed from the top

The correct option is C 1.5
Apparent depth of coin as seen from top = $\frac{x}{μ_{1}} + \frac{x}{μ_{2}} = x$
$\Rightarrow \frac{x}{μ_{1}} + \frac{x}{μ_{2}} = 1 \Rightarrow \frac{1}{3} + \frac{1}{μ} = 1 \Rightarrow μ = 1.5$