1.

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 :-

Answer»

`(1-(1)/(mu_1))h_2+(1-(1)/(mu_2))h_1`
`(1+(1)/(mu_2))h_1+(1+(1)/(mu_1))h_2`
`(1-(1)/(mu_1))h_1+(1-(1)/(mu_2))h_2`
`(1+(1)/(mu_1))h_2-(1+(1)/(mu_2))h_1`

SOLUTION :
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`


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