A beaker contains water up to a height `h_(1)` and kerosene of height `h_(2)` above water so that the total height of (water + kerosene) is `(h_(1) + h_(2))` . Refractive index of water is `mu_(1)` and that of kerosene is `mu_(2)` . The apparent shift in the position of the bottom of the beaker when viewed from above is :-

Options

A. `(1-(1)/(mu_1))h_2+(1-(1)/(mu_2))h_1`
B. `(1+(1)/(mu_2))h_1+(1+(1)/(mu_1))h_2`
C. `(1-(1)/(mu_1))h_1+(1-(1)/(mu_2))h_2`
D. `(1+(1)/(mu_1))h_2-(1+(1)/(mu_2))h_1`

Answer

A, B, C

Explanation

Apparent depth `=("Real depth")/("Refractive index")`
Shift of bottom `x=d_o-d_i`
`=d_o-(d_o)/(mu)`
`=(1-(1)/(mu))d_o` …. (1)
For water `x_1=(1-(1)/(mu_1))h_1` and
For kerosene `x_2=(1-(1)/(mu_2))h_2`
`therefore` Total shift of bottom
`=x_1+x_2`
`=(1-(1)/(mu_1))h_1+(1-(1)/(mu_2))h_2`