calculate the image distance for an object of height 12mm from a concave lens of focal lenght 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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