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  • A beaker contains water up to a height and kerosene of height <img alt="\mathrm{2\ h}" src=

A beaker contains water up to a height \mathrm{1\ h} and kerosene of height \mathrm{2\ h} above water so that the total height of (water + kerosene) is \mathrm{ (h_{1}+h_{2})} . Refractive index of water is \mathrm{ \mu _{1}} and that of kerosene is \mathrm{ \mu _{2}} . The apparent shift in the position of the bottom of the beaker when viewed from above is:

Option: 1

\mathrm{\left(1-\frac{1}{\mu_1}\right) h_2+\left(1-\frac{1}{\mu_2}\right) h_1}


Option: 2

\mathrm{\left(1+\frac{1}{\mu_1}\right) h_1-\left(1+\frac{1}{\mu_2}\right) h_2}


Option: 3

\left(1-\frac{1}{\mu_1}\right) \mathrm{h}_1+\left(1-\frac{1}{\mu_2}\right) \mathrm{h}_2


Option: 4

\left(1+\frac{1}{\mu_1}\right) \mathrm{h}_2-\left(1+\frac{1}{\mu_2}\right) \mathrm{h}_1


Answers (1)

best_answer


Apparent shift of bottom position of beaker in water is \mathrm{\mathrm{x}_1=\mathrm{h}_1-\frac{\mathrm{h}_1}{\mu_1}=\mathrm{h}_1\left(1-\frac{1}{\mu_1}\right)}
Apparent shift of bottom position in kerosene is \mathrm{\mathrm{x}_2=\mathrm{h}_2-\frac{\mathrm{h}_2}{\mu_2}=\mathrm{h}_2\left(1-\frac{1}{\mu_2}\right)}
Total shift =\mathrm{x}_1+\mathrm{x}_2=\mathrm{h}_1\left(1-\frac{1}{\mu_1}\right)+\mathrm{h}_2\left(1-\frac{1}{\mu_2}\right)

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Riya

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