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A 20 cm thick glass slab of refractive index 1.5 is kept in front of a plane mirror. Find the position of the image (relative to mirror) as seen by an observer through the glass slab when a point object is kept in air at a distance of 40 cm from the mirror. |
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Answer» Solution :As THICKNESS of glass slab is 20 cm and its `MU = (3//2)`, so shift produced by it will be `x=d[1-(1)/(mu)]=20[1-(2)/(3)]=(20)/(3) cm` (towards the mirror)  The image of object in the mirror is `(100)/(3)`cm from the mirror. The glass slab shift this image towards the observer by `(20)/(3) cm` Therefore the position of the final image from the 100 20 80 mirror w.r.t observer is `(100)/(3) -(20)/(3) =(80)/(3) cm` |
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