Question

if an object of 7cm height is placed at a distance of 12 cm form a convex lens of focal length 8cm find the position,nature and height of the image?

Answer 1

Given

$h_o$ = Height of object = 7 cm

u = Object distance = 12 cm

f = Focal length of convex lens = 8 cm

To find

Position, nature and height of the image

Solution

v = Image distance

From the lens maker formula we have

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}\\\Rightarrow \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\Rightarrow \dfrac{1}{v}=\dfrac{u-f}{uf}\\\Rightarrow v=\dfrac{uf}{u-f}\\\Rightarrow v=\dfrac{12\times 8}{12-8}\\\Rightarrow v=24\ \text{cm}$

Since the sign is positive the image is on the other side 24 cm from the lens and it is real in nature. This means the image is formed at 2F.

Magnification is given by

$M=\dfrac{h_i}{h_o}=-\dfrac{v}{u}\\\Rightarrow \dfrac{h_i}{h_o}=-\dfrac{v}{u}\\\Rightarrow h_i=-h_o\dfrac{v}{u}\\\Rightarrow h_i=-7\times \dfrac{24}{12}\\\Rightarrow h_i=-14\ \text{cm}$

The height of the image is 14 cm. Since it is negative it is inverted.

Answer 2

Answer:

according to my

findings:

v=24cm

hi= -14 cm

Explanation:

you may check attached pages for more clarification

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