Question

An object $10cm$ tall is placed on the principal axis of a concave lens of focal length $10cm$ at a distance of $15cm$ from it. Find the position and the nature of the image formed by the lens and draw the ray diagram.

Answer 1

Step 1: Given data and the figure:

The ray diagram is,

Given,

1. Height of object($h_o$)=$10cm$
2. Focal length($f$)=$-10cm$ [negative sign indicated that the lens is concave]
3. Distance of object($u$)=$-15cm$ [negative sign indicated that the object is on the left side of the lens]

Formula used

1. lens formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ [$v$=distance of image from the lens]
2. magnification, $m=\frac{v}{u}=\frac{h_o}{h_i}$ [$h_i$=height of image]

Step 2: Finding the image distance:

Now, finding $v$,

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \frac{1}{-10}=\frac{1}{v}-\frac{1}{-15} \\ \frac{1}{v}=\frac{1}{-10}+\frac{1}{-15}=-\frac{25}{150}=-\frac{1}{6} \\ \Rightarrow v=-6cm \end{gathered}$$

Step 3: Finding the nature of image:

Now, finding $h_i$,

$$\begin{gathered} m=\frac{v}{u}=\frac{h_o}{h_i} \\ \Rightarrow \frac{-6}{-15}=\frac{10}{h_i} \\ \Rightarrow h_i=4cm \end{gathered}$$

Thus, the nature of image formed ,

1. virtual, erect and
2. diminished because its height is less than the height of the object