Question

An upright object is placed at a distance of 40 cm in front of a convergent lens of a focal length of 20 cm. A convergent mirror of focal length 10 cm is placed at a distance of 60 cm on the other side of the lens. The position and size of the final image will be ______.

Options

• 20 cm from the convergent mirror, the same size as the object.

• 40 cm from the convergent mirror, the same size as the object.

• 40 cm from the convergent lens, twice the size of the object.

• 20 cm from the convergent mirror, twice the size of the object.

Answer 1

An upright object is placed at a distance of 40 cm in front of a convergent lens of a focal length of 20 cm. A convergent mirror of focal length 10 cm is placed at a distance of 60 cm on the other side of the lens. The position and size of the final image will be 40 cm from the convergent mirror, the same size as the object.

Explanation:

Given: The focal length of the lens is f₁ = 20 cm, the focal length of the mirror is f₂ = -10 cm, and the distance between the mirror and the lens is d = 60 cm. The item is u₁ = 40 cm away from the lens.

To find: The final image's position and magnification.

The image from a convex lens is located (v₁) and magnified (m₁) as follows: 

`1/v_1 = 1/f_1 - 1/u_1 = 1/20 - 1/40 = 1/40, v_1 = 40  cm, m_1 = v_1/u_1 = 1.`

The first image, formed at v₁ = 40 cm behind the lens, is the same size as the object but inverted (I₁), and it serves as an object for a concave mirror.

For the second illustration: 

`u_2 = -20` cm, `f_2 = -10` cm.

Image location from a concave mirror:

`1/v_2 = 1/f_2 - 1/u_2 = 1/(-10) - 1/(-20) = -1/20,`

`v_2 = -20` cm, `m_2 = v_2/u_2 = 1.`

So, at v₂ = -20 cm, the second image is formed in front of the concave mirror, which is the same size as an object but inverted (I₂), and serves as an object for the lens.

u₃ = 40 cm, f₁ = 20 cm for image three (final image I₃).

The final image's location:

`1/v_3 = 1/f_1 - 1/u_3 = 1/20 - 1/40 = 1/40,`

`v_3 = 40` cm, `m_3 = u_3/v_3 = 1`.

As a result, the final image is formed at v₃ = +40 cm in front of the lens, in the same position as the object but inverted. The image has the same dimensions as the object.