QJEEM19Ph7945: JEE Main 2019 Physics - Class 12 JEEM19Ph7945 Solution: An upright object is placed at a distance of 40 cm in front of…

In given system of lens and mirror, position of object O in front of lens is at a distance 2f. i.e. u = 2f = 40 cm

So, image (I₁) formed is real, inverted and at a distance, v = 2f = 2 $$\times$$ 20 = 40 cm, (behind lens) magnification, $$m_1=\frac{v}{u}=\frac{40}{40}=1$$

Thus, size of image is same as that of that of object. This image (I₁) acts like a real object for mirror.

As object distance for mirror is

u = C = 2f = $$-$$ 20 cm

where, C = centre of curvature.

So, image (I₂) formed by mirror is at 2f.

$$\therefore$$ For mirror v = 2f = 2($$-$$ 10) = $$-$$ 20 cm

Magnification, $$m_2=-\frac{v}{u}=-\frac{(-20)}{(-20)}=-1$$

Thus, image size is same as that of object.

The image I₂ formed by the mirror will act like an object for lens.

As the object is at 2f distance from lens, so image (I₃) will be formed at a distance 2f or 40 cm. Thus, magnification,

$$m_3=\frac{v}{u}=\frac{40}{40}=1$$

So, final magnification, $$m=m_1\times m_2\times m_3=-1$$

Hence, final image (I₃) is real, inverted of same size as that of object and coinciding with object.