Question

an object 50 cm tall is placed on the principal axis of a convex lens it is 20 CM tall images formed on the screen placed at a distance of 20 cm from the lens can create the focal length of the lens​

Answer 1

Given

• Height of object (ho) = +50cm
• Height of image (hi) = -20cm
• Object distance (u) = -20cm
• Focal length (f) = ?

Note :

Height of object is bigger than height of image . That's why image is formed real and inverted

As we know that

Magnification = It is the ratio of height of image or size of image to the height or size of object.

$\implies\sf m=\dfrac{v}{u}=\dfrac{h_i}{h_o} \\ \\ \\ \implies\sf \dfrac{v}{u}=\dfrac{h_i}{h_o} \\ \\ \\ \implies\sf \dfrac{-v}{20}=\dfrac{-20}{50} \\ \\ \\ \implies\sf \dfrac{-v}{20}=\dfrac{-2}{5} \\ \\ \\ \implies\sf 5v=40 \\ \\ \\ \implies\sf v=\dfrac{40}{5}=+8cm$

Now, apply lens formula

$\implies\sf \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{1}{8}-\dfrac{(-1)}{20}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{5+2}{40}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{1}{f}=\dfrac{7}{40} \\ \\ \\ \implies\sf f=+5.7cm$

Hence, the focal length of convex lens is + 5.7cm

Answer 2

GIVEN

Height of the object (h$\sf{_o}$) = 50 cm

Height of the image (h$\sf{_i}$) = 20 cm

Image distance (v) = 20 cm

$\underline{\rule{200}3}$

TO FIND

The focal length (f) of the lens.

$\underline{\rule{200}3}$

WE NEED TO KNOW

$\bigstar\ \sf{Magnification=\dfrac{h_i}{h_o}=\dfrac{v}{u} }$

$\bigstar$ Lens Formula :-

$\sf{\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} }$

$\underline{\rule{200}3}$

SOLUTION

$\sf m=\dfrac{v}{u}=\dfrac{h_i}{h_o} \\ \\ \longrightarrow\sf \dfrac{v}{u}=\dfrac{h_i}{h_o} \\ \\ \longrightarrow\sf \dfrac{20}{u}=\dfrac{-20}{50} \\ \\ \longrightarrow\sf \dfrac{1}{u}=\dfrac{-1}{5} \\ \\ \longrightarrow\sf{-u=5}\\\\\boxed{\sf{\longrightarrow u=-5\ cm}}$

$\rule{180}2$

Now,

Applying the lens formula :-

$\sf{\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} }$

$\sf{\longrightarrow \dfrac{1}{20}-\dfrac{1}{-5}=\dfrac{1}{f} }\\\\\sf{\longrightarrow \dfrac{1}{20}+\dfrac{1}{5}=\dfrac{1}{f} }\\\\\sf{\longrightarrow \dfrac{1}{f} =\dfrac{1+4}{20} }\\\\\sf{\longrightarrow \dfrac{1}{f} =\dfrac{5}{20} }\\\\\sf{\longrightarrow f=\dfrac{1}{4} }\\\\\boxed{\sf{\longrightarrow f=4\ cm }}$

Therefore, the focal length (f) of the convex lens is 4 cm.