Question

To find the focal length of a convex mirror, a student records the following data:
Object pin:$22.2\;cm$
Convex lens:$32.2\;cm$
Convex mirror:$45.8\;cm$
Image pin:$71.2\;cm$
The focal length of the convex lens is $f_{1}$ and that of the mirror is $f_{2}$. Then taking index correction to be negligibly small, $f_{1}$ and $f_{2}$ are close to:
\[\begin{align}
  & A.{{f}_{1}}=7.8cm,{{f}_{2}}=12.7cm \\
 & B.{{f}_{1}}=12.7cm,{{f}_{2}}=7.8cm \\
 & C.{{f}_{1}}=7.8cm,{{f}_{2}}=25.4cm \\
 & D.{{f}_{1}}=15.6cm,{{f}_{2}}=25.4cm \\
\end{align}\]

Answer 1

Hint: Using the lens and mirror formula, we can solve this sum. Here we have an object and an image. Considering the object produces a virtual image, we can use it in turn as a virtual object to the final image.
Formula used:
$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$ and $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Complete answer:
The mirror formula is the relationship between the distance of an object $u_{m}$, distance of image $v_{m}$ and the focal length of the lens $f_{m}$. This law can be used for both concave and convex mirrors with appropriate sign conventions. Given as $\dfrac{1}{u_{m}}+\dfrac{1}{v_{m}}=\dfrac{1}{f_{m}}$
Similarly, lens formula is the relationship between the distance of an object $u_{l}$, distance of Image $v_{l}$ and the focal length of the lens $f_{l}$. This law can be used for both concave and convex lens with appropriate sign conventions. Given as $\dfrac{1}{v_{l}}-\dfrac{1}{u_{l}}=\dfrac{1}{f_{l}}$
Since the lens in kept immediately after the pin, we have, $v_{l}=71.2-32.2=39cm$ is the virtual image and the pin at $u_{l}=32.2-22.2=10cm$ then, we have $\dfrac{1}{39}-\dfrac{1}{10}=\dfrac{1}{f_{l}}$
$\implies f_{1}=7.8cm$
Hence from the given option, option A or option C is correct.

Similarly, for the mirror, we have $v_{m}=71.2-45.8=25.4cm$ and $u_{m}=48.5-32.2-39=-22.7$
Substituting, we have, $\dfrac{1}{25.4}+\dfrac{1}{-22.7}=\dfrac{1}{f_{m}}$
$\implies \dfrac{1}{25.4}+\dfrac{1}{22.7}=\dfrac{1}{f_{m}}$
$\implies0.0393 +0.0440=0.0833=\dfrac{1}{f_{m}}$
$\implies f_{m}=12cm$
Since we have rounded off the values, we are getting $f_{m}=12cm$ and $f_{1}=7.8cm$
Since in the given options, option has a value of $f_{2}$ close to $f_{m}=12cm$

So, the correct answer is “Option A”.

Note:
The thickness of the lens is neglected. The formula can be used for any lens and when the object is placed is anywhere on the principal axis. Both the lens and the mirror formula can be used for both concave and convex mirrors. However, we must use the appropriate sign conventions.