Question

An object of height 4.0 cm is placed at a distance of 30 cm from the optical centre ‘O’ of a convex

lens of focal length 20 cm. Draw a ray diagram to find the position and size of the image formed.

Mark optical centre ‘O’ and principal focus ‘F ’ on the diagram. Also find the approximate ratio of

size of the image to the size of the object.​

Answer 1

Answer:

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Answer 2

$\orange{\mathfrak{Given}} \\ \\ \tt Object\: distance \:(u) = -30 \:cm \\ \tt Focal\:length\:(f) = 20\:cm \\ \tt Object\:height\:({h}_{o}) = 4\:cm \\ \\ \orange{\mathfrak{To\:find}} \\ \\ \tt i.Position\:of\:image \\ \tt ii.Size\:of\:image \\ \tt iii.Ratio\:of\:size\:of\:image\:to\:the\:size\:of\:the\:object \\ \\ \tt \orange{\mathfrak{Solution}} \\ \\ \tt We\:know\:that \\ \\ \tt \boxed{\bold{\underline{\green{Lens\:Formula: \frac{1}{v} - \frac{1}{u} = \frac{1}{f} }}}} \\ \\ \tt \implies \frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \\ \\ \tt \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{30} \\ \\ \tt \implies \frac{1}{v} = \frac{3 -2}{60} \\ \\ \tt \implies \frac{1}{v} = \frac{1}{60} \\ \\ \tt \boxed{\bold{\underline{\red{\tt v = 60\:cm}}}} \\ \\ \tt We\:know\:that \\ \\ \tt \boxed{\bold{\underline{\green{\tt Magnification = \frac{{h}_{i}}{{h}_{o}} = \frac{v}{u} }}}} \\ \\ \tt \implies \frac{{h}_{i}}{4} = \frac{60}{30} \\ \\ \tt \implies {h}_{i} = 8\:cm \\ \\ \tt \boxed{\bold{\underline{\red{\tt {h}_{i} = 8\:cm}}}} \\ \\ \tt ratio = \frac{{h}_{i}}{{h}_{o}} \\ \\ \tt ratio = \frac{8}{4} \\ \\ \tt \boxed{\bold{\underline{\red{ratio=2\:cm}}}}$