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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



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F = 20cm, u = -28cm

 \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \\ \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}=70cm

m= \frac{v}{u} = \frac{70}{-28} =-2.5

The image is formed at 70cm to the right side of lens. It is real, inverted  and magnified.
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