Question

An object of height 4cm is placed before a convex lens of focal length 10 cm on principal axis at a distance of 15cm . Find the image distance, size of image and write characteristics of image

Answer 1

$\frac{1}{f }= \frac{1}{v} - \frac{1}{u}$
$\frac{1}{f} + \frac{1}{u} = \frac{1}{v}$
$= > \frac{1}{10} - \frac{1}{15} = \frac{1}{v}$
$= > \frac{3 - 2}{30} = \frac{1}{v}$
$= > \frac{1}{30} = \frac{1}{v}$
$= > v = 30cm$
$\\ \\ = > \frac{v}{u} = \frac{h1}{h}$
$= > \frac{30}{ - 10} = \frac{h1}{4}$

$= > - 3 \times 4 = h1$
$= > h1 = - 12cm$
Image formed is enlarged erect and real.
With having size greater than object.
Image formed is between beyond C.

Answer 2

$\huge\mathfrak\red{Ello\: Mate!!!}$

Ur AnSWeR iS in Attachment✔✔

height of image=12cm

image distance=30cm✌

Nature øF image:-
➡Image formed is real and inverted
➡ size of image is larger then the size of object .
➡Image formed is beyond C.

Hope that will heLp uhh ØuT❤