Question

a 5 cm tall object is placed on priciple of axis of a convex lens of focal length 50 cm at distance of 40 cm from it find the natural position and size of the image

Answer 1

Given

• height of object , h₁ = 5 cm
• focal length of convex lens, f = +50 cm
• position of object , u = - 40 cm

To find

• Nature of image
• position of image , v = ?
• size of image , h₂ = ?

Formulae Used

→ formula for magnification of lens

$\boxed{\sf{m=\dfrac{v}{u}=\dfrac{h_2}{h_1}}}$

→ Lens formula

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

( where m = magnification of lens , v = position of image , u = position of object , h₂ = height of image , h₁ = height of object , f is the focal length of lens )

Solution

Calculating position of image using Lens formula

$\longmapsto\sf{\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}\\\\ \\\longmapsto \sf{\dfrac{1}{50}=\dfrac{1}{v}-\dfrac{1}{(-40)}}\\\\\\\longmapsto\sf{\dfrac{1}{50}=\dfrac{1}{v}+\dfrac{1}{40}}\\\\\\\longmapsto\sf{\dfrac{1}{v}=\dfrac{1}{50}-\dfrac{1}{40}}\\\\\\\longmapsto\sf{\dfrac{1}{v}=\dfrac{4-5}{200}}\\\\\\\longmapsto\boxed{\boxed{\sf{v=-200\;cm}}}\orange{\bigstar}$

Calculating height of image using formula for magnification

$\longmapsto\sf{m=\dfrac{v}{u}=\dfrac{h_2}{h_1}}\\\\\\\longmapsto\sf{\dfrac{v}{u}=\dfrac{h_2}{h_1}}\\\\\\\longmapsto\sf{\dfrac{-200}{-40}=\dfrac{h_2}{5}}\\\\\\\longmapsto\sf{h_2=5\times 5}\\\\\\\longmapsto\boxed{\boxed{\sf{h_2=25\;cm}}} \orange{\bigstar}$

$\rule{202}2$

● Position of image is at 200 cm in front of the lens.

(In v = -200 cm , negative sign shows that image is formed in front of the lens and 200 cm shows the magnitude)

● Size of image is 25 cm. (that is 5 times magnified image)

( since, h₂ = 25 cm and h₁ = 5 cm hence, we

can say that image is 5 times magnified )

● Nature of image is Virtual and Erect .

( In v = -200 cm for convex lens , negative sign shows that image formed is virtual and In h₂ = 25 cm positive height of image shows that image image formed is erect. )

$\rule{202}2$