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Calculate the image distance for an object of height 12mm at a distance of 0.20m from a concave lens of focal length 0.30m ,and state the nature and size

of the image



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Object distance = u = -0.2m = -20cm
focal length = f = -0.3m = -30cm
image distance = v 

 \frac{1}{v} - \frac{1}{u} = \frac{1}{f}  \\  \\ \frac{1}{v} - \frac{1}{-20} = \frac{1}{-30} \\  \\ \frac{1}{v} = \frac{1}{-30} + \frac{1}{-20} = \frac{-3-2}{60} = \frac{-5}{60}  \\  \\ v=- \frac{60}{5} =-12cm

m= \frac{v}{u} = \frac{-12}{-20}= \frac{3}{5}   \\  \\ m= \frac{h'}{h}  \\  \\  \frac{3}{5} = \frac{h'}{12mm}  \\  \\ h'= \frac{12*3}{5}=7.2mm,\ erect,\ virtual
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