The apparent vertical shift of the image of a coin placed at the bottom of a container filled with a liquid having depth d and refractive index ' $μ$ ' is _____

μ

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(μ - \frac{1}{d})

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d (1 - \frac{1}{μ})

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d (μ + 1)

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Solution

The correct option is C $d (1 - \frac{1}{μ})$
Since we know $μ = \frac{r e a l d e p t h}{a p p a r e n t d e p t h}$
hence, $a p p a r e n t d e p t h = \frac{r e a l d e p t h}{μ}$
hence $v e r t i c a l s h i f t = d - \frac{d}{μ}$
$δ = d (1 - \frac{1}{μ})$

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The apparent vertical shift of the image of a coin placed at the bottom of a container filled with a liquid having depth d and refractive index 'μ' is _____

The correct option is C d(1−1μ)Since we know μ=realdepthapparentdepthhence, apparentdepth=realdepthμhence verticalshift=d−dμδ=d(1−1μ)

The apparent vertical shift of the image of a coin placed at the bottom of a container filled with a liquid having depth d and refractive index ' $μ$ ' is _____

The correct option is C $d (1 - \frac{1}{μ})$
Since we know $μ = \frac{r e a l d e p t h}{a p p a r e n t d e p t h}$
hence, $a p p a r e n t d e p t h = \frac{r e a l d e p t h}{μ}$
hence $v e r t i c a l s h i f t = d - \frac{d}{μ}$
$δ = d (1 - \frac{1}{μ})$