Question

a 1 cm tall object is placed perpendicular to the principal axis of a convex lens of a focal length 20cm the distance of the object from the lens is 15 cm find the nature position size and magnification of the image

Answer 1

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Answer 2

Answer: The height of image is 4 cm and the image is virtual, erect.

Explanation:

Given that,

Height of the object h = 1 cm

Focal length f = 20 cm

Distance of the object u = -15 cm

Using lens formula,

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

$\dfrac{1}{20}-\dfrac{1}{15}=\dfrac{1}{v}$

$\dfrac{1}{v} =\dfrac{1}{-60}$

$v = -60 cm$

Therefore, the image will formed at -60 cm from the lens

We know that,

The magnification is

$m = \dfrac{v}{u}$

$m = \dfrac{60}{15} = 4$

The magnification is 4.

$m = \dfrac{h'}{h}$

$4 = \dfrac{h'}{1}$

$h' = 4 cm$

Hence, The height of image is 4 cm and the image is virtual, erect.