Question

An object $1.4cm$high when placed in front of a convex lens at a distance of$6cm$, forms a virtual image at a distance of$24cm$ from the lens. Calculate (a) the focal length of the lens (b) the size of the image.

Answer 1

Step 1: Given data

height of object$\left(h_o\right)=1.4cm$

height of image$\left(h_i\right)=?$

distance of the object from the lens$\left(u\right)=-6cm$ (object distance is taken as negative)

distance of the image from the lens$\left(v\right)=-24cm$ (the image is virtual)

focal length$\left(f\right)=?$

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Step 2: (a)calculating the focal length

Using the lens formula

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{f}=\frac{1}{-24}-\frac{1}{(-6)} \\ \Rightarrow \frac{1}{f}=\frac{1}{-24}+\frac{1}{6}=\frac{-1+4}{24}=\frac{3}{24}=\frac{1}{8} \\ \Rightarrow f=8cm \end{gathered}$$

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Step 3:(b) calculating the size of the image

Using the expression,

$$\begin{gathered} \frac{h_i}{h_o}=\frac{v}{u} \\ \Rightarrow \frac{h_i}{1.4}=\frac{-24}{-6} \\ \Rightarrow h_i=\frac{1.4\times 24}{6}=5.6cm \end{gathered}$$

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The focal length of the lens is $8cm$and the size of the image is $5.6cm$.