A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of - (a) 20 cm from the plane mirror, it would be a virtual image. (b) 20 cm from the lens, it would be a real image. (c) 30 cm from the lens, it would be a real image. (d) 30 cm from the plane mirror, it would be a virtual image.

Answer: (a). Lens: u = -60, f = 30: v = 60 cm, i.e. 20 cm behind the mirror (virtual object for the mirror). Plane mirror forms its image 20 cm in front of the mirror, i.e. 20 cm right of the lens. Light returning through the lens: object 20 cm away, f = 30: 1/v = 1/30 - 1/20 ⇒ v = -60 cm, i.e. on the mirror side, virtual. This is 60 - 40 = 20 cm behind the plane mirror. So the answer is (a) 20 cm from the plane mirror, virtual.

A point object is placed at a distance of 60 cm from a convex lens of… | NEET 2021

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Combination of Lenses

A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of -

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Answer: (a)

Lens: $u = -60$ , $f = 30$ : $v = 60\text{ cm}$ , i.e. 20 cm behind the mirror (virtual object for the mirror).
Plane mirror forms its image 20 cm in front of the mirror, i.e. 20 cm right of the lens.
Light returning through the lens: object 20 cm away, $f = 30$ : $\frac{1}{v} = \frac{1}{30} - \frac{1}{20} \Rightarrow v = -60\text{ cm}$ , i.e. on the mirror side, virtual.
This is $60 - 40 = 20\text{ cm}$ behind the plane mirror.
So the answer is (a) 20 cm from the plane mirror, virtual.

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