Question 227 of 295

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

A$40cm$ from the convergent lens, twice the size of the objectB$20cm$ from the convergent mirror, twice the size of the objectC$40cm$ from the convergent lens, same size of the objectD$20cm$ from the convergent mirror, same size of the object

Show answer and solution▾

Answer: (C) $40cm$ from the convergent lens, same size of the object

First refraction, $u=-40cm; f=+20 cm$
From lens formula, $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{1}{v}-\dfrac{1}{-40}=\dfrac{1}{20}\Rightarrow \dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{40}$
$v=+40 cm$
${m}_{1}=-1$
First reflection, $u=-20 cm;f=-10 cm$
From mirror formula, $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{1}{v}-\dfrac{1}{20}=\dfrac{1}{-10}$
$\dfrac{1}{v}=\dfrac{1}{-10}+\dfrac{1}{20}\Rightarrow v=-20 cm$
${m}_{2}=-1$
Second refraction, $u=-40 cm, f=+20 cm$
From lens formula, $\dfrac{1}{v}-\dfrac{1}{-40}=\dfrac{1}{20}\Rightarrow v=+40cm$
${m}_{3}=-1$
${m}_{total}={m}_{1}\times {m}_{2}\times {m}_{3}=-1$
Finally, image will be at distance $40cm$ from lens and is of same size as object.