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Question

# A glass slab of referactive index μ=1.5 and thickness of 15 cm is placed in front of a convex mirror as shown in figure. The distance between the actual position of pole and image of object O as seen by the observer will be:

A
9 cm
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B
25 cm
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C
15 cm
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D
18 cm
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Solution

## The correct option is C 15 cmThe distance of object from pole of mirror is : u=−(20+15+10)=−45 cm However, due to presence of glass slab, the position of object will shift in direction of incident ray. S=t(1−1μ)=15(1−11.5) ⇒S=5 cm ∴ u=−(45−5)=−40 cm f=+40 cm Applying mirror formula, 1v+1u=1f ⇒1v+1−40=140 or 1v=240 ∴ v=+20 cm i.e. image is formed at a distance 20 cm behind mirror. Once again during journey of reflected ray, slab will produce a shift 'S' in direction of reflected ray. ⇒ Image will get shifted by S=5 cm towards the observer. Now image position as observed by observer is : v′=+(20−5)=+15 cm behind the actual pole of the mirror. ∴ Distance between the actual position of pole of the mirror and image as seen by the observer is 15 cm. Hence, option (c) is the correct answer. Why this question? Caution: When reflected rays pass through slab, shift will be produced in direction of reflected rays. Hence, image will appear to come closer by 'S' for observer.

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