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 object starts to move with speed of `12 cm//sec` towards the slab then find the speed of image as seen by observer [given `n_(3)=` air =(ref. index `=1`)].

Options

A. `16 cm//sec`
B. `9 cm//sec`
C. `12 cm//sec`
D. none of these

Answer

A

Explanation

`d'=(d_(1))/(1//4//3)+(d_(2))/(9//8)+d_(3)`
`v'=(3v_(1))/(4)+0+0 rArr v' =(3)/(4) xx 12 =9 cm//sec`.