A tank contains two different liquids which do not mix with each other. The lower and upper liquid are at depth `h_(1)` and `h_(2)` respectively and of refractive indices `mu_(1)` and `mu_(2)`. An object `O` is located at the bottom, when see vertically from above. Locate the position of image of the object `O` as seen from above.

Options

A. `(h_(1))/(mu_(1))-(h_(2))/(mu_(2))`
B. `(h_(1))/(mu_(1))+(h_(2))/(mu_(2))`
C. `(h_(1))/(mu_(1))+(h_(2))/(mu_(1))`
D. `(h_(1))/(mu_(2))-(h_(2))/(mu_(1))`

Explanation

(b) Let `/_BOP=/_i,/_BDP=/_r,/_CEQ=/_i'`

Refractrion at `P`
`mu_(1)sin i=mu_(2) sin r`
`(mu_(1))/(mu_(2))=(sin r)/(sin i)=(tan r)/(tan i)=(h_(1))/(BD)`
`implies BD=(h_(1)mu_(2))/(mu_(1))`
At `Q,mu_(2) sin r =sin i'`
`mu_(1)=(sin i')/(sin r)+(tan i')/(tan r)=(BD+BC)/(CE)`
`implies CE=(DB+BC)/(mu_(2))=(h_(1))/(m_(1))+(h_(2))/(mu_(2))`