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A 5cm tall object is placed on the principal axis of a convex lens of fl 50 cm at a distance of 40 cm from it .use lens formula to find the nature and

position of the image



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H=5cm, u = -40cm, f = 50cm

 \frac{1}{v} - \frac{1}{u} = \frac{1}{f}  \\  \\  \frac{1}{v} = \frac{1}{u} + \frac{1}{f} =\frac{1}{-40} + \frac{1}{50} \\  \\  \frac{1}{v}= \frac{-5+4}{200} = \frac{-1}{200}  \\  \\ v=-200cm

m= \frac{h'}{h}= \frac{v}{u}  = \frac{-200}{-40} =5 \\  \\ h'=5* 5cm=25cm

The image is formed at 200cm to the left from lens, erect and virtual.
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