Question

An object 3 cm high is placed perpendicular to the principal axis of a concave lens of focal length 15 cm. The image is formed at a distance of 10 cm from the lens. Calculate a. The distance of object placed b. size and nature of the image formed.

Answer 1

(a)

Lens formula:


$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \frac{1}{-15}=\frac{1}{-10}-\frac{1}{u} \\ \frac{1}{u}=\frac{1}{-10}+\frac{1}{15}=\frac{-5}{150} \\ u=-30cm \end{gathered}$$

Hence, the distance of the object is 30 cm.

(b)

Magnification:

$$\begin{gathered} m=\frac{v}{u} \\ =\left(\frac{-10}{-30}\right) \\ =\frac{1}{3} \end{gathered}$$

Size of the image:

$$\begin{gathered} h_i=m{h}_o \\ =\frac{1}{3}\left(3\right) \\ =1cm \end{gathered}$$

Here, the positive magnification implies erect and virtual image.

As the magnification is less than 1, the image is smaller than the object.