Question

50 points ...,.......





An object is kept at a distance of 0.2 m from a convex lens of 0.15m, find the position of the image and its nature.

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Answer 1

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heya , here is ur answer..

$\bold{\red{\underline{Given:-}}}$

$\mathrm{u=-0.2\:m=-20\:cm}$

$\mathrm{f=0.15\:m=15\:cm}$

$\bold{\red{\underline{To find:-}}}$ $\mathrm{Position\:and\:nature\:of\:image}$

$\bold{\red{\underline{Solution:-}}}$

By using lens formula-

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \\ \frac{1}{15} = \frac{1}{v} - \frac{1}{( - 20)}$

$\frac{1}{15} = \frac{1}{v} + \frac{1}{20} \\ \\ \frac{1}{15} - \frac{1}{20} = \frac{1}{v}$

$\frac{4 - 3}{60} = \frac{1}{v} \\ \\ \frac{1}{60} = \frac{1}{v} \\ \\ 60\: cm = v$

$\mathrm$

Since, the value of "v" is positive, therefore the nature of the image formed is real and inverted.

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Answer 2

Answer in attachment ...

given :

u = -0.2 m = -20 cm

f = 0.15 m = 15 cm

60 cm position of the image and real and inverted its nature.