Answers

2016-03-29T20:23:40+05:30
Radius of curvature, R = + 3.00 m;
Object-distance, u = – 5.00 m;
Image-distance, ν = ?
Height of the image, h′= ?

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

Since 1/ν + 1/u = 1/f

or, 1/ν = 1/f - 1/u = + 1 / 1.50 - 1 / (-5.00) = 1 / 1.50 + 1/ 5.00 

= (5.00 + 1.50) / 7.50

ν = +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; = - 1.15 m / -5.00 m
= +0.23
The image is virtual, erect and smaller in size by a factor of 0.23.
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