There is a slab of refractive index `n_(2)` placed as shown. Medium on two side are `n_(1)` and `n_(3)`. Width of slab is `d_(2)`. An object is placed at distance `d_(1)` from surface `AB` and observer is at distance `d_(3)` from surface `CD`. Given `n_(1)=` air [ref. index `=1`] and `n_(2)` is glass [ref. index `= 3//2`], `d_(1) = 12 cm` and `d_(2) = 9 cm,d_(3) = 4 cm`.

If `n_(3)= (4)/(4)` then distance of image of object seen by oberver is :

Options

A. 25 cm
B. 30 cm
C. 28 cm
D. none of these

Answer

C

Explanation

`d'=(d_(1))/(n_(1)//n_(3))+(d_(2))/(n_(2)//n_(3))+d_(3)`
`=(12)/(1//4//3)+(9)/(3//2//4//3) +(4)/(1)`
`= 12 xx(4)/(3)+(8)/(1)+4 =16+8 +4=28 cm`.