Question

A 4.5 CM needle is placed at 12 CM away from a convex mirror of focal length 15cm. Give the location of the image and the magnification. describe what happen as the needle is moved further from the mirror.

Answer 1

ANSWERS :-


Here,

object size , h1 = 4.5 cm

object distance, u = –12 cm

focal length, f = + 15 cm

image distance , v = ?

magnification, m = ?

As. 1/u + 1/v = 1/f

1/v = 1/f – 1/u

Now,

putting u = –12 cm and f = +15cm , we get

1/v = 1/15 - 1/-12 = 4+5/60 = 9/60

v = 60/9 = 6.7 cm

✴image is formed 6.7 CM behind the convex mirror. it must be virtual and erect.


if h2 is size of image , then m = h2/h1 = -v/u

or,

m = h2/h1 = –(6.7)/–12 = 0.558

h2 = 0.558

h1 = 0.558 * 4.5 = 2.5 cm

⚡As the needle is moved further from the mirror, image moves away from the mirror till it is at focus F of the mirror the size of the image goes on decreasing.

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❤BE BRAINLY ❤
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Answer 2

Answer:

Explanation:

Given :-

u = - 12 cm

f = + 15 cm

Solution :-

Using lens formula,

1/f = 1/u + 1/v

⇒ 1/v = 1/f - 1/u

⇒ 1/v = 1/15 - 1/- 12

⇒ 1/v = 60/9

⇒ v = 6.7 cm

Magnification, m = - (- v/u) = h₂/h₁

⇒ m = - 6.7/- 12

⇒ m = 0.558

m = h₂/h₁

h₂ = h₁ × m

⇒ h₂ = 0.558 × 4.5

⇒ h₂ = 2.5 cm.

As the needle is moved further from the mirror, image moves to the focus and the size of image goes on decreasing.