Question

An object of height 3 cm is placed in front of a convex lens of focal length 10 cm at a ditance of 15 cm.
(a) what is the distance of the image formed?
(b) what is the nature of the image?
(c) what is the height of the image?

Answer 1

lens formula problem

Answer 2

(a) Image distance, v = 30 cm

(b) Image is inverted.

(c) height of the image is 6 cm.

Explanation:

It is given that,

Height of the object, h = 3 cm

Object distance, u = -15 cm

Focal length of the convex lens, f = +10 cm (for convex lens it is positive)

(a) Let the object is placed at a distance of v from the lens. It can be calculated using lens formula as :

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

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

$\dfrac{1}{v}=\dfrac{1}{10}+\dfrac{1}{(-15)}$

v = 30 cm

So, the image will be formed at a distance of 30 cm from the lens.

(b) Magnification,

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

h' is the size of image

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

$h'=3\times \dfrac{30}{-15}$

h' = -6 cm

So, the size of image is 6 cm and it is inverted.

Hence, this is the required solution.

Learn more,

Mirror's formula

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