(C) The apparent depth $d'$ of the bottom of the container when viewed from above through multiple layers of liquids is given by the formula:$d' = \fr

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(C) The apparent depth $d'$ of the bottom of the container when viewed from above through multiple layers of liquids is given by the formula:$d' = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$Here,$h_1 = 60 \ cm$,$\mu_1 = 1.2$,$h_2 = H$,and $\mu_2 = 1.6$.So,$d' = \frac{60}{1.2} + \frac{H}{1.6} = 50 + \frac{H}{1.6}$.The actual depth of the bottom is $d = 60 + H$.The apparent shift is given by the difference between the actual depth and the apparent depth:$\text{Shift} = d - d'$$40 = (60 + H) - (50 + \frac{H}{1.6})$$40 = 10 + H - \frac{H}{1.6}$$30 = H(1 - \frac{1}{1.6})$$30 = H(\frac{1.6 - 1}{1.6}) = H(\frac{0.6}{1.6})$$30 = H(\frac{6}{16}) = H(\frac{3}{8})$$H = 30 \times \frac{8}{3} = 80 \ cm$.

Class 12 Physics Ray Optics and Optical Instruments Refraction through Plane Surface and Glass Slab

Medium

$A$ container contains a liquid with refractive index of $1.2$ up to a height of $60 \ cm$ and another liquid having refractive index $1.6$ is added to height $H$ above the first liquid. If viewed from above,the apparent shift in the position of the bottom of the container is $40 \ cm$. The value of $H$ is . . . . . . $\ cm$. (Consider liquids are immiscible)

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A
$20$
B
$30$
C
$80$
D
$10$

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Ray Optics and Optical Instruments

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Refraction through Plane Surface and Glass Slab

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$A$ coin is placed at the bottom of an empty beaker and is focused by a microscope. When water is poured into the beaker to a height of $10 \, cm$,how much must the microscope be moved to focus it again?

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