Question

Plz solve this question

Answer 1

GIVEN :-

●Focal length of the lens = 20 cm

●Object distance = 40cm (from the lens)

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●From right angled triangle ABC :-

$= > tan \alpha = \frac{AB}{BC} \\ \\ = > tan \alpha = \frac{5}{40} \\ \\ = > \alpha = {tan}^{ - 1} ( \frac{5}{40} )$
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●From right angle triangle ECD :-

$= > tan \alpha = \frac{DE}{DC} \\ \\ = > tan \alpha = \frac{h}{20} \: \: \: \: ..(DB - BC = DC = 20) \\ \\ putting \: the \: value \: of \: angle \: \alpha \\ \\ = > tan( {tan}^{ - 1} ( \frac{5}{40} )) = \frac{h}{20} \\ \\ = > \frac{5}{40} = \frac{h}{20} \\ \\ = > h = 2.5cm$
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●so your answer is option "b"

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_-_-_-_☆$BE \: \: \: \: BRAINLY$☆_-_-_-_

Answer 2

@ Heya... Monali...@
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I'm going to give short way to solve these kinds of questions because of shortage of time in many such exams....for more details you can see the answer of Deepsbhargav......His explaination is too brilliant.............

Have a look for your answer dear, in the above pic:-----

Hope it will help you!
Thanks!!
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