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Question

# An object is placed 21 cm in front of a concave mirror of radius of curvature 10 cm. A glass slab of thickness 3 cm and μ=1.5 is then placed close to the mirror in the space between the object and the mirror. The position of final image formed is

A
3.94cm
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B
4.3cm
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C
4.93cm
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D
3.94cm
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Solution

## The correct option is C −4.93cmDue to glass slab, increase in path=(μ−1)t{μ=re fractive index t= thickness of slab } torease path =(1.5−1)3t=1.5 cmnew initial distance =u′=−(21+1.5)=−22.5 cm Now, 1−22.5+1v=1−5⇒1v=122.5−15⇒1v=5−22.522.5×5∴v=22.5×5−17.5=−6.43 cm⇒∨=−6.43 cm Geometrical distance =−(6.43−1.5)=−4.93 cm

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