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 refractive index 1.5 is then placed close to the mirror in the space between the object and the mirror. The distance of the near surface of the slab from the mirror is 1 cm. The final image from the mirror will be formed at :

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

Answer 1

The correct option is D 7.67 cm

The shift in position of object due to slab is given by

$\Delta t=t(1-\frac{1}{\mu })=3(1-\frac{1}{\frac{3}{2}})=3(1-\frac{2}{3})=1cm$

The apparent distance of object from mirror $21-1=20cm$

The position of image formed by concave mirror can be found by mirror formula

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$...(i)

As object and focus of mirror lies left side of pole of mirror

$u=-20,f=-\frac{R}{2}=-5$

substituting values in equation (i)

$\frac{1}{v}+\frac{1}{-20}=\frac{1}{-5}$

$\frac{1}{v}=\frac{1}{20}-\frac{4}{20}$

$v=-\frac{20}{3}$

As image is other side of the slab,the image will be again shifted by 1 cm away from mirror.

Final distance of image from mirror$=\frac{20}{3}+1cm=7.67cm$