Question

A convex lens of focal length 20cm is placed coaxially with a concave mirror of focal length 10cm at a distance of 50cm 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.

Answer 1

The ray comes from infinity$\left( {u = \infty } \right)$ and is incident on the convex lens$\left( {f = + 20cm} \right)$. We use the lens formula to find the position of 1st image v formed. This v serves as the object for the concave mirror($f = - 10cm)$. Then we use a mirror formula to find the position of the image formed.

Complete step by step answer: Given that the ray comes from infinity$\left( {u = \infty } \right)$and is incident on the convex lens$\left( {f = + 20cm} \right)$. Using lens formula,
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\\$
Substituting the value of u and f we get,
$\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{20}} + \dfrac{1}{\infty } \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{20}} + 0 = \dfrac{1}{{20}} \\\ \therefore v = 20cm \\\$
This v serves as the object for the concave mirror($f = - 10cm)$.
Using mirror formula,
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
Substituting the value of u$=$distance between lens and mirror-distance of image from the optical centre of the lens.
$u = - (50 - 20) = - 30cm$(negative because object is infront of the mirror, using conventions)
Given focal length$f = - 10cm$.
$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{ - 10}} - \dfrac{1}{{ - 30}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{30}} - \dfrac{1}{{10}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{{ - 2}}{{30}} = \dfrac{{ - 1}}{{15}} \\\ \therefore v = - 15cm \\\$
v is negative which shows the image is formed in front of the mirror.
This v again serves as an object for the lens. u$=$distance between lens and mirror-distance of image from the pole of the mirror.
$u = - (50 - 15) = - 35cm$ [negative due to sign conventions]
To find the position of image formed we use lens formula,
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\\$
Substituting the value of u and f we get,

$$\begin{aligned} \dfrac{1}{v} &= \dfrac{1}{f} + \dfrac{1}{u} \\\ \Rightarrow \dfrac{1}{v} &= \dfrac{1}{{20}} + \dfrac{1}{{ - 35}} \\\ \Rightarrow \dfrac{1}{v} &= \dfrac{1}{{20}} - \dfrac{1}{{35}} = \dfrac{3}{{140}} \\\ \therefore v &= \dfrac{{140}}{3}cm = 46.67cm \\\ \end{aligned}$$

Therefore the final image is formed 46.67cm to the left of the lens.

Note: Sign conventions should be kept in mind to avoid mistakes in sign used. Distance is always measured from the centre of the lens or mirror. Distance in the direction of incident light is taken as positive and opposite to direction of incident light is taken as negative.