Question

A lens of focal length 15cm forms an erect image three the size of the object .what is the distance between the object and image ​

Answer 1

Answer:

YR ANS..

Explanation:

MAKE MY ANS BRAINLIEST

Answer 2

Explanation:

ANSWER

In△PAQ

Sumoftwooppositeinteriorangles=Exteriorangle

α=θ+∠θ−(i)

In△PAR

α+θ+∠R=180

∘

α=180

∘

−θ−∠R−(ii)

In△PMQ

∠MPQ=180−∠Q−90

∘

=90−∠Q

∠APM=∠APQ−∠MPQ

=θ−(90−∠Q)

=θ+∠Q−90

∘

−(iii)

∵(i)=(ii)

=>θ+∠Q=180−θ−∠R

=>2θ=180−∠Q−∠R

=>θ=90−

2

1

(∠Q+∠R)

puttingthisvalueofθin(iii)

∠APM=∠Q+90−

2

1

(∠Q+∠R)−90

=

2

∠Q

−

2

∠R

=

2

1

(∠Q−∠R)

∴a=2