# Using the formula and sign convention for refraction by spherical lens .find the position nature and magnification of the image formed by convex lens of focal length 20 cm when object is at a distance of 28 cm from lens

Given

F = 20cm, u = -28cm

using the lens formula

\begin{aligned} &\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ &\begin{array}{l} \frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{1}{-28}+\frac{1}{20} \\ \frac{1}{v}=\frac{-5+7}{140}=\frac{2}{140} \\ v=\frac{140}{2}=70 \mathrm{cm} \\ m=\frac{v}{u}=\frac{70}{-28}=-2.5 \end{array} \end{aligned}

The image is formed at 70cm to the right side of the lens.

and   It is real, inverted, and magnified.

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