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

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5.67 cm

7.67 cm

4.67 cm

6.67 cm

answer is D.

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Detailed Solution

$u = 21 cm, f = \frac{R}{2} = \frac{10}{2} = 5 cm$

On introducing the glass slab, the object as well as the image will be shifted from the mirror through a distance

$d = t (1 - \frac{1}{μ}) = 3 (1 - \frac{1}{1 .5}) = 1 cm$

so that apparent distance of the object = 20 cm i.e., u = 20 cm

By the mirror formula, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$v = - \frac{20}{3} cm = - 6 .67 cm$

Distance of the final image from the mirror

= 6.67 + 1 = 7.67 cm

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