Question

A concave lens has focal length of 20 cm. At what distance from the lens, a 5 cm tall object be placed so taht the forms an image at 15 cm from the lens? What calculate the size of the image formed.​

Answer 1

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Answer 2

Given :-

• Focal length of concave lens, f = -20 cm

(focal length of concave lens is always negative)

• height of object , h₁ = 5 cm
• position of image , v = -15 cm

(concave lens always form an image in front of lens hence, v is negative)

To find :-

• size of image , h₂ = ?

Formula required :-

• Lens formula

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

• Formula for magnification of lens

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

(where f is focal length of lens , u is position of object ,v is position of image , h₂ is height of image and h₁ is height of object)

Solution :-

Let, position of object = u

and height of image = h₂

then,

Using Lens formula

$\implies\sf{\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}\\\\\implies\sf{\dfrac{1}{-20}=\dfrac{1}{-15}-\dfrac{1}{u}}\\\\\implies\sf{\dfrac{1}{u}=\dfrac{1}{-15}+\dfrac{1}{20}}\\\\\implies\sf{\dfrac{1}{u}=\dfrac{-4+3}{60}}\\\\\implies\boxed{\sf{u=-60\;cm}}$

Now,

using formula for magnification of Lens

$\implies\sf{m=\dfrac{v}{u}=\dfrac{h_2}{h_1}}\\\\\implies\sf{\dfrac{v}{u}=\dfrac{h_2}{h_1}}\\\\\implies\sf{\dfrac{-15}{-60}=\dfrac{h_2}{5}}\\\\\implies\sf{h_2=\dfrac{1}{4}\times 5}\\\\\implies\underbrace{\underline{\boxed{\sf{h_2=1.25\;cm}}}}\red{\bigstar}$

Hence,

size of image formed will be 1.25 cm .