Question
Mon May 28, 2012 By:

Wed May 30, 2012
Due to the presence of plate, effective distance of the object is shifted towards the mirror by an amount

OO1= t(1-(1/mu))
= 6(1- (1/1.5)
= 2cm

Hence effective object distance is u= (42-2)=40 cm
According the new sign convention, u=-40cm

f=-r/2 = -16/2 =8 cm
from the mirror formula we have,

1/f =1/v +1/u
1/8 =1/v - 1/40
-1/v = -1/40 - 1/8
1/v = 1/40 + 1/8
1/v= 6/40
v = 40/6 = 6.66 cm

If the slab would not have been there, the image would be at I1, where PI1= 6.66 cm, but since the slab is present, the light ray while returning back passes through the slab and gets displaced by 2 cm away from the slab. Thus the image will actually at I, where PI = 6.66 +2 = 8.66 cm
look at the picture below

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