Question

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Answer 1

Answer = 0.23

⬇️⬇️Explanation ⬇️⬇️

Radius of curvature, R = +3.00 m;

Object - distance, u = - 5.00 m;

Image - distance, v = ?

Height of the image, h'=?

Focal length, f = R/2 = + 3.00 m / 2 = + 1.50 m

Since

1/v + 1/u = 1/f ( mirror formula)

Or,

1/v = 1/f - 1/u

= + 1 / 1.50 - 1/(-5.00)

= 1/1.50 + 1/5.00

=( 5.00 + 1.50) / 7.50

v = + 7.50/6.50

= + 1.15 m

The image is 1.15 m at the back of the mirror.

Magnification, m = h' / h

= - v/u = 1.15m / - 5.00 m

= + 0.23

The image is virtual, errect and smaller in size by a factor of 0.23

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Answer 2

Answer:

Hey mate your answer is 0.23