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

Options

A. ``4.67cm
B. 6.67cm
C. 5.67cm
D. 7.67cm

Answer

D

Explanation

`u=21cm, f=(R)/(2)=(10)/(2)=5cm`
On introducing the glass slab, the object as well as the image will be shifted from the mirror through a distance
`d=t(1-(1)/(mu))=3(1-(1)/(1.5))=1cm`, so that aparent distance of the object `=20cm ` i.e., `u=20cm`
By the mirror formula, `(1)/(v)+(1)/(u)=(1)/(f)`
`v=-(20)/(3)cm=-6.67cm`
Distance of the final image from the mirror `=6.67+1`
`=7.67cm`.