Question

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Answer 1

Hii its tom85

let the size of object be x then size of image 3x

m = -v/u = 3x/x= 3/1

-v/u= 3

v=-30

using lense formula

1/v - 1/u = 1/f

1/f = u - v /uv

f= 300 /10 + 30

f= 15 /2

now we know that on combination of two lens there power must be added to give a unique focal (power) length

p1 + p2 = pt

p1 = 1/ focal length of concave lens

p2= ?

pt= 1/combine focal length

2/15 +1/30 = pt

ft= 30/5 = 6cm convex lens

__________/\__________☺️

hope it may helps you

Answer 2

SOLUTION

Focal length of combination of lens=feq:

$= > \frac{1}{feq} = \frac{1}{f1} + \frac{1}{f2} \\ = > m = - 3 \\ = > \frac{v}{u} = - 3 = v = - 3u \\ = > \frac{1}{feq} = \frac{1}{v} - \frac{1}{u} \\ = > \frac{1}{feq} = \frac{1}{ - 3u} - \frac{1}{u} \\ = > \frac{1}{feq} = - \frac{4}{ 3u} \\ = > \frac{1}{feq} = \frac{ - 4}{3( - 10)} \\ = > \frac{1}{feq} = \frac{30}{4} \\ \\ = > \frac{1}{feq} = \frac{1}{f1} + \frac{1}{f2} \\ = > \frac{4}{30} = \frac{1}{ - 30} + \frac{1}{f2} \\ = > \frac{1}{f2} = \frac{1}{30} + \frac{4}{30} \\ = > \frac{1}{f2} = \frac{5}{60} = \frac{1}{6} \\ \\ = > f2 = 6cm$

Option (3)✓

Convex lens

hope it helps ☺️