An object is placed at a distance of $\displaystyle 30$ cm in front of a convex lens of focal length $\displaystyle 15$ cm. Use lens formula to determine the position of the image. What will be the size of the image in this case ?

\(\displaystyle u=-30 \mathrm{~cm}, f=15 \mathrm{~cm}\) \[\begin{aligned} & \frac{1}{\mathrm{v}}-\frac{1}{u}=\frac{1}{f} \\ & \frac{1}{\mathrm{v}}=\frac{1}{f}+\frac{1}{u} \\ & \frac{1}{\mathrm{v}}=\frac{1}{15}+\frac{1}{-30} \\ & \mathrm{v}=30 \mathrm{~cm} \end{aligned} \]

\(\displaystyle \mathrm{m}=\frac{v}{u}=+\frac{30}{-30}=-1\) Hence, size of image will be same as the object. / Size of image will be same as size of object because when object is kept at \(\displaystyle 2 \mathrm{~F}_{1}\), image will be formed at \(\displaystyle 2 \mathrm{~F}_{2}\).