Question

An upright object is placed at a distance of 40 cm in front of a convergent lens of focal length 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
A). 40 cm from the convergent mirror, the same size as the object.
B). 20 cm from the convergent mirror, the same size as the object.
C). 20 cm from the convergent mirror, twice the size as the object.
D). 40 cm from the convergent mirror, twice the size as the object.

Answer 1

Hint: In order to determine the position and size of the final image, we break down the entire phenomena that occur here into individual processes, i.e. refraction, reflection, and refraction respectively. We apply the formulae for each process and derive the answer.

Formulae used:
Lens formula: $\dfrac{1}{{\text{f}}} = \dfrac{1}{{\text{v}}} - \dfrac{1}{{\text{u}}}$
Magnification = $\dfrac{{\text{v}}}{{\text{u}}}$
Here f is the focal length, v is the height of the image, and u is the height of the object.

Complete step-by-step solution:
We make an appropriate figure of the given phenomena which looks like:

Here ${{\text{I}}_1}$ is the image formed due to refraction from the lens w.r.t the object O, ${{\text{I}}_2}$ is the image formed due to reflection from the mirror for which the object is ${{\text{I}}_1}$ and ${{\text{I}}_3}$ is the image formed due to the refraction from the lens once again for which the object is ${{\text{I}}_2}$.
The final image is formed after these 3 individual processes,
Refraction of O from lens forming ${{\text{I}}_1}$
We apply the lens formula, here u = -40 cm and f = 20 cm
(The object distance is negative because it is measured in the opposite direction of all the other distances)
Hence, $
  \dfrac{1}{{\text{f}}} = \dfrac{1}{{\text{v}}} - \dfrac{1}{{\text{u}}} \Rightarrow \dfrac{1}{{20}} = \dfrac{1}{{\text{v}}} - \dfrac{1}{{ - 40}} \Rightarrow {\text{v = 40cm}} \\
  {\text{m = }}\dfrac{{\text{v}}}{{\text{u}}} = \dfrac{{40}}{{ - 40}} = - 1 \\
$

Reflection of ${{\text{I}}_1}$ from the mirror forming ${{\text{I}}_2}$
We apply the lens formula, here u = 20 cm and f = -10 cm
Hence, $
  \dfrac{1}{{\text{f}}} = \dfrac{1}{{\text{v}}} + \dfrac{1}{{\text{u}}} \Rightarrow \dfrac{1}{{ - 10}} = \dfrac{1}{{\text{v}}} - \dfrac{1}{{20}} \Rightarrow {\text{v = - 20cm}} \\
  {\text{m = }}\dfrac{{\text{-v}}}{{\text{u}}} = \dfrac{{ -(-20)}}{{-20}} = - 1 \\
$
Refraction of ${{\text{I}}_2}$ from the lens forming ${{\text{I}}_3}$
We apply the lens formula, here u = -40 cm and f = 20 cm
(The object distance is negative because it is measured in the opposite direction of all the other distances)
Hence, $ \dfrac{1}{{\text{f}}} = \dfrac{1}{{\text{v}}} - \dfrac{1}{{\text{u}}} \Rightarrow \dfrac{1}{{20}} = \dfrac{1}{{\text{v}}} - \dfrac{1}{{ - 40}} \Rightarrow {\text{v = 40cm}} \\
  {\text{m = }}\dfrac{{\text{v}}}{{\text{u}}} = \dfrac{{40}}{{ - 40}} = - 1 $
Hence the final image is at a distance of 20 cm from the convergent mirror.
Now the total magnification of the original object is given by the product of all the individual magnifications of images formed in each reflection or refraction, i.e.
Total magnification m = $-1 \times -1 \times -1 = -1$.

Hence the final image is 20 cm from the convergent mirror and the same size as the object.
Option B is the correct answer.

Note: In order to answer questions of this type the key is to know the definitions of convergent lens and mirror and the phenomena of reflection and refraction and their respective formulae.
Converging lens: a convex lens in which light rays that enter it parallel to its axis converge at a single point on the opposite side
A concave mirror, or converging mirror, has a reflecting surface that bulges inward (away from the incident light). Concave mirrors reflect light inward to one focal point.
Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated.
Refraction is the change in direction of a wave passing from one medium to another or from a gradual change in the medium.