A glass slab of thickness 3 cm and refractive index 3/2 is placed on ink mark on a piece of paper. For a looking at the mark at a distance 5.0 cm above it, distance of the mark will appear to be

3.0 cm

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4.0 cm

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4.5

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5.0 cm

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Solution

The correct option is A 4.0 cm
The apparent depth of link marked
$= \frac{r e a l d e p t h}{μ} = \frac{3}{3 / 2} = 2 c m$
Thus person views mark at a distance $= 2 + 2 = 4 c m$

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A glass slab of thickness 3 cm and refractive index 3/2 is placed on ink mark on a piece of paper. For a looking at the mark at a distance 5.0 cm above it, distance of the mark will appear to be

The correct option is A 4.0 cmThe apparent depth of link marked=real depthμ=33/2=2 cmThus person views mark at a distance =2+2=4 cm

The correct option is A 4.0 cm
The apparent depth of link marked
$= \frac{r e a l d e p t h}{μ} = \frac{3}{3 / 2} = 2 c m$
Thus person views mark at a distance $= 2 + 2 = 4 c m$