A vessel of height 2d is half- filled with a liquid of refractive index `sqrt 2` and the other half with a liquid of refractive index `n`.(The given liquids are immiscible).Then the apparent depth of the inner surface of the bottom of the vessel (neglecting the thickness of the bottom of the vessel) will be

Options

A. `(n)/(d(n + sqrt2))`
B. `(d(n + sqrt2))/(n sqrt2)`
C. `(sqrt2n)/(d(n + sqrt2)`
D. `(nd)/(d + sqrt2n)`

Answer

B

Explanation

`mu = (Real depth)/(Apparent depth) = (d)/(x)`
`therefore` Due to `1^(st)` liquid, `sqrt2 = (d)/(x_(1))` or `x_(1) = (d)/(sqrt2)`
Due to the second liquid, `n = (d)/(x_(2))`
`therefore x_(2) = (d)/(n)`
`therefore` Total apparent depth = `x_(1) + x_(2) = (d)/(sqrt2) + (d)/(n)`
Total apparent depth = `(d(n+ sqrt2))/(n sqrt2)`