A beaker contains water up to a height h_{1} | vedclass

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(C) The apparent shift produced by a medium of thickness $h$ and refractive index $\mu$ is given by the formula: $\Delta h = h \left( 1 - \frac{1}{\mu

Vedclass · Accepted Answer

$\left( 1 - \frac{1}{\mu_{1}} \right) h_{1} + \left( 1 - \frac{1}{\mu_{2}} \right) h_{2}$

Vedclass · Answer

(C) The apparent shift produced by a medium of thickness $h$ and refractive index $\mu$ is given by the formula: $\Delta h = h \left( 1 - \frac{1}{\mu} \right)$.In this problem,there are two layers: water of height $h_{1}$ with refractive index $\mu_{1}$ and kerosene of height $h_{2}$ with refractive index $\mu_{2}$.The total apparent shift is the sum of the shifts produced by each individual layer:Shift due to water = $h_{1} \left( 1 - \frac{1}{\mu_{1}} \right)$.Shift due to kerosene = $h_{2} \left( 1 - \frac{1}{\mu_{2}} \right)$.Therefore,the total apparent shift = $h_{1} \left( 1 - \frac{1}{\mu_{1}} \right) + h_{2} \left( 1 - \frac{1}{\mu_{2}} \right)$.

Similar Questions

Assume the situation shown in the figure. Water $(\mu_w = 4/3)$ is filled in a beaker up to a height of $10 \, cm$. $A$ plane mirror is fixed at a height of $5 \, cm$ above the water surface. The distance of the image of the object $O$ from the mirror is ...... $cm$.

$A$ fish at a depth of $12 \, cm$ in water is viewed by an observer on the bank of a lake. To what height is the image of the fish raised? (Refractive index of water $\mu = 4/3$)

Three immiscible liquids of densities $d_1 > d_2 > d_3$ and refractive indices $\mu_1 > \mu_2 > \mu_3$ are put in a beaker. The height of each liquid column is $\frac{h}{3}$. $A$ dot is made at the bottom of the beaker. For near normal vision,find the apparent depth of the dot.

$A$ rectangular tank of depth $8 \, m$ is full of water $(\mu = 4/3)$. The bottom is seen at an apparent depth of:

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