Question

A lens forms a real image 3 cm high of an object 1 cm high. If the separation of object and image is 15 cm, find the focal length of the lens.

Answer 1

Given,
Height of image (​h') = - 3 cm (negative because image is real)
Height of ​object = 1 cm
Let us first define the sign convention:
Since the object distance is always negative, let it be -u.
Since the focal length is positive for a convex lens, let it be +f .

Magnification =$\frac{{\mathrm{h}}^{\text{'}}}{\mathrm{h}}$ =$\frac{\mathrm{v}}{\mathrm{u}}$
$\frac{\mathrm{v}}{\mathrm{u}}=-3$

∴​ $\mathrm{v}=-3\mathrm{u}$

(i) Since object distance (u) is always negative, and the ratio $\frac{\mathrm{v}}{\mathrm{u}}$ here is negative, v must be positive, i.e., it is on the right side of the lens.

Now, we have object distance + image distance = 15

v + (-u) = 15 (using sign convention)
$-3\mathrm{u}-\mathrm{u}=15$

-$$\begin{gathered} -4\mathrm{u}=15 \\ \mathrm{u}=\frac{-15}{4} \\ Applyingthel\mathrm{ens}\mathrm{formula}: \\ \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\ \frac{1}{\mathrm{f}}=\frac{1}{-3\mathrm{u}}-\frac{1}{\mathrm{u}} \\ \frac{1}{\mathrm{f}}=\frac{-4}{3\mathrm{u}} \\ \mathrm{f}=\frac{3\mathrm{u}}{-4}=\frac{3\times {\displaystyle \frac{-15}{4}}}{4}=\frac{45}{16}=2.81\mathrm{cm} \\ H\mathrm{ence},the\mathrm{focal}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{lens}\mathrm{is}+2.81\mathrm{cm}. \end{gathered}$$