Question

A convex lens of focal length 20 cm is placed coaxially with a concave mirror of focal length 10 cm at a distance of 50 cm apart from each other. A beam of light coming parallel to the principal axis is incident on the convex lens. Find the position of the final image formed by this combination. Draw the ray diagram showing the formation of the image

Answer 1

Let us first locate the image of the point object S formed by the convex lens.

Here:

u = -∞ cm and f = 20 cm

From the lens formula, we have:

${\displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f}}$

${\displaystyle \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}}$

${\displaystyle \Rightarrow \frac{1}{v}=\frac{1}{20}+\frac{1}{-\infty }}$

${\displaystyle \Rightarrow \frac{1}{v}=\frac{1}{20}}$

⇒ v = 20 cm

The positive sign shows that the image is formed to the right of the lens, as shown in the following figure.

The image I₁ is formed in front of the mirror and hence, acts as a real source for the mirror. The concave mirror forms the image I₂, whose distance from the mirror can be calculated as:

${\displaystyle \frac{1}{v}=\frac{1}{u}=\frac{1}{f}}$

Here:

u=-30 cm ; f=-10 cm

${\displaystyle \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}}$

${\displaystyle \Rightarrow \frac{1}{v}=-\frac{1}{10}+\frac{1}{30}}$

${\displaystyle \Rightarrow \frac{1}{v}=\frac{1-3}{30}=\frac{-2}{30}}$

⇒ v = − 15 cm

Hence, the final image is formed at a distance of 15 cm from the concave mirror, as shown in the following figure.