Question

of
pla
An object height 6cm is placed
perpendicular to the principal axis of
a concave lens of focal length 5cm.
Determine the position, size and nature
of the image of of the distance of th
object from the lens is 10 cm.​

Answer 1

Answer:-

Given : u=−10 cm f=−5 cm ho=6cm

$\frac{1}{v} = \frac{1}{u} + \frac{1}{f}$

$\frac{1}{v} = \frac{1}{ - 10} + \frac{1}{ - 5}$

$\frac{1}{v} = \frac{ - 1 - 2}{10}$

$\frac{1}{v} = \frac{ - 3}{10}$

$v = \frac{ - 10}{3}$

$v = - 3.33cm$

Magnification (M) :- v/u

$\frac{v}{u} = \frac{ - 10}{3} \times ( \frac{ -1}{10} ) = \frac{1}{3}$

Also, magnification M = v/u= h2/h1

$\frac{h2}{h1} = \frac{v}{u}$

$\frac{h2}{6} = \frac{1}{3}$

$h2 = \frac{6}{3}$

$h2 = 2cm$

position:- behind the mirror

size:- diminished

nature of image =virtual and erect

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