A glass slab $(μ = 1.5)$ of the thichness 2 cm is placed on a spot. The shift of spot if it is viewed from top is

\frac{2}{3} c m

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\frac{4}{3} c m

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\frac{1}{3} c m

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\frac{5}{3} c m

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Solution

The correct option is A $\frac{2}{3} c m$
$Apparent depth = \frac{Real Depth}{μ} t^{'} = \frac{2}{μ} = \frac{20}{15} = \frac{4}{3} t^{'} = \frac{4}{3} Δ t = t - t^{'} = 2 - \frac{4}{3} = \frac{6 - 4}{3} = \frac{2}{3} c m$

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A glass slab (μ=1.5) of the thichness 2 cm is placed on a spot. The shift of spot if it is viewed from top is

The correct option is A 23cmApparent depth =Real Depthμt′=2μ=2015=43t′=43Δt=t−t′=2−43=6−43=23 cm

A glass slab $(μ = 1.5)$ of the thichness 2 cm is placed on a spot. The shift of spot if it is viewed from top is

The correct option is A $\frac{2}{3} c m$
$Apparent depth = \frac{Real Depth}{μ} t^{'} = \frac{2}{μ} = \frac{20}{15} = \frac{4}{3} t^{'} = \frac{4}{3} Δ t = t - t^{'} = 2 - \frac{4}{3} = \frac{6 - 4}{3} = \frac{2}{3} c m$