Question

An object of 4 cm height is held 30 cm away from a concave lens of focal length 20 cm find the position nature and the size of the image formed

Answer 1

Explanation:

highly diminished

virtual and erect image

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Answer 2

Explanation:

Given that,

Height of the object, h = 4 cm

Object distance, u = -30 cm

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

We need to find the position nature and the size of the image formed. Let v is the image distance. It is given by :

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

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{(-20)}+\dfrac{1}{(-30)}$

v = -12 cm

So, the image is formed in front of concave lens at a distance of 12 cm. Let h' is the size of image. So,

$\dfrac{h'}{h}=\dfrac{v}{u}$

$\dfrac{h'}{4}=\dfrac{-12}{-30}$

h' = 1.6 cm

So, the size of the image is 1.6 cm. Hence, this is the required solution

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Mirror's formula

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