An object is placed 21 cm in front of a concave mirror of radius of curvature 10cm. A glass slabe of thickness 3cm and refractive index 1.5 is then placed close to the irror in the space between the object and the mirror. The distance of the near surface of the slabe from the mirror is 1cm. The final image from the mirror will be formed at

Answer

C

Explanation

The rays origination from A ( the point object) suffer refraction before striking the concave mirror. For the mirror the rays are coming from `A^``
such that `AA^``=shift=t(1-(1)/(mu))`
Therefore the object distance
`u=OA^`=OA-A A^`=21-t(1-(1)/(mu))`
`=21-3(1-(1)/(1.5))=20cm`
`:'``v=(uf)/(u-f)=(20xx5)/(20-5)=(20)/(3) cm=6.67cm`
The reflected rays again pass through the glass slab. The image should have formed at B is the absence of glass slab. But. due to its presence the image is formed at `B^``.
Therefore image distance`=OB+BB^``
`(20)/(3)+t(1-(1)/(mu))`,(20)/(3)+1=(23)/(3)=7.67`cm`
(##JMA_RWO_C16_108_S02##)