Question

An object is placed 21 cm infront of a concave mirror of radius of curvature 20 cm. A glass slab of thickness 3 cm and refractive index 1.5 is placed close to the mirror in the space between object and the mirror. The distance of the nearer surface of the slab from the mirror is 10 cm. Find the position of the final image formed

• A. Final image coinsides with object ie. V=21 cm
• B. Final image coinsides with object ie V= 30 cm
• C. Final image coinsides with object ie V=20 cm
• D. Final image coinsides with object ie V=25 cm

Answer 1

The correct option is C Final image coinsides with object ie V=20 cm

Focal length (f) = $\frac{r}{2}=\frac{20}{2}=10cm$

Due to the glass slab, the object appears closer than the original distance by a distance d

$d=\left(1-\frac{1}{\mu }\right)\times t$

$d=\left(1-\frac{1}{1.5}\right)\times 3$

$d=\frac{1}{3}\times 3=1cm$

Hence object distance will be

$u=21-1=20cm$

Using mirror formula,

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{-10}=\frac{1}{-20}+\frac{1}{v}$

$\frac{1}{-10}+\frac{1}{20}=\frac{1}{v}$

$\frac{1}{v}=-\frac{1}{20}$

$v=-20cm$

The object distance and the image distance are same. So they both will coincide and v = 20 cm