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There is a slab of refractive index n(2)...

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`)].

A

`8 cm//sec`

B

`180 cm//sec`

C

`12 cm//sec`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
Apparent distance (t)
`= (d_(1))/(n_(1)//n_(3))+(d_(2))/(n_(2)//n_(3))+d_(3)`
but `n_(1)=n_(3)=1` & `n_(2)=3//2`
`d=(d_(1))/(1)+(d_(2))/(3//2) +d_(3)`
but `d_(2)` and `d_(3)` are constant when only object is moving
So `d' = d'_(1)+0+0`
`rArr v_(1) = v_(0) =12 cm//sec`.
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