Question

4. An object 3 cm high is placed at a distance of 10 cm from a concave mirror which
produces a virtual image of length 4.5 cm.
(1) What is the magnification produced?
(i) Is magnification negative or positive?
(m) What are the values of u and v for the mirror?​

Answer 1

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Answer 2

Explanation:

Hey!

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Height of the object (h) = 3 cm

Distance of object from concave mirror (u) = -8 cm

Height of the virtual image formed (h') = 4.5

Let's find 'v' first =

- v/u = h'/h

v = - u × h'/h

v = - ( -8) × 4.5 / 3

v = 12 cm

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(i) Focal length of mirror = ?

1/f = 1/v + 1/u

1/f = 1/12 + 1/ -8

1/f = -1/24

1 × 24 = -f

24 = -f

f = -24 cm

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(ii) Position of image (v) = 12 cm

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(iii) Ray diagram - Refer pic

(Image is vertical , thus erect)

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Hope it helps...!!!