Question

An object is first placed at $24cm$ from the lens and then at a distance of $16cm$ from the lens. If the magnification of the image formed in both the cases is same, determine the focal length of the lens.

Answer 1

Step 1: Finding the magnification of the first case:

Taking the first case, when object distance, $u=-24cm$ (negative sign implies that the object is on the left side)

Using the lens formula to find image distance $\left(v\right)$,

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{f}=\frac{1}{v}-\frac{1}{-24} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{-24} \\ \Rightarrow \frac{1}{v}=\frac{24-f}{24f} \\ \Rightarrow v=\frac{24f}{24-f} \end{gathered}$$

Finding the magnification,

$m=\frac{v}{u}=\frac{\frac{24f}{24-f}}{24}=\frac{f}{24-f}$

Step 2: Finding the magnification of the second case:

Taking the second case, when object distance, $u=-16cm$ (negative sign implies that the object is on the left side) , here this case will make virtual image $\Rightarrow v=-v$

Using the lens formula to find image distance $\left(v\right)$,

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{f}=\frac{1}{-v}-\frac{1}{-16} \\ \Rightarrow \frac{1}{-v}=\frac{1}{f}+\frac{1}{-16} \\ \Rightarrow \frac{1}{-v}=\frac{16-f}{16f} \\ \Rightarrow -v=\frac{16f}{16-f} \\ \Rightarrow v=\frac{16f}{f-16} \end{gathered}$$

Finding the magnification,

$M=\frac{v}{u}=\frac{\frac{16f}{f-16}}{16}=\frac{f}{f-16}$

Step 3: Finding the focal length:

According to the question, $m=M$

$$\begin{gathered} \frac{f}{24-f}=\frac{f}{f-16} \\ \Rightarrow 24-f=f-16 \\ \Rightarrow 2f=40 \\ \Rightarrow f=20cm \end{gathered}$$

Thus, the focal length is $20cm$.