Question

An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Answer 1

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Size of the object, h1 = 3 cm Object distance, u = −14 cm Focal length of the concave lens, f = −21 cm Image distance = v According to the lens formula, we have the relation: Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual. The magnification of the image is given as: Hence, the height of the image is 1.8 cm. If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.

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Answer 2

Image will form at a distance of 8.4 cm from the lens and it is virtual and erect.

Explanation:

Given that,

Size of object, h = 3 cm

Object distance, u = -14 cm

Focal length of the concave lens, f = -21 cm

The image produced is given by :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(-21)}+\dfrac{1}{(-14)}\\\\v=-8.4\ cm$

Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{-8.4}{-14}\\\\m=0.6$

It shows image is virtual and erect.

Learn more,

Lens formula

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