An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm .find the position of image ,its nature and size

Dear Student

Given, Distance of the object from the convex mirror (u) = -20 cm
Radius of Curvature : 30 cm
Focal length ( f) = (30 / 2) cm
= 15 cm

Now, as we know,
1/u + 1/v = 1/f [ where f = focal length , v = distance of the image from the mirror and u = distance of the object from the mirror]

 1/(-20) + 1/v = 1/15
 1/v = 1/15 - (-1/20)
 1/v = 1/15 + 1/20
 1/v = (4 + 3) / 60
1/v = 7/60
 v = 60/7
 v = 8.57 cm

Therefore, distance of the image from the mirror is 8.57cm.

Now, Height of the object ( H_o) = 5 cm
Distance of the image from the mirror ( v) = 8.5cm
Distance of the object from the mirror ( u) = -20 cm
Since,Magnification (m) = H_i/H_o    [ where H_o = height of the object and H_i = height of the image]

M = -v/u = - 8.5 cm / -20 cm = +0.42
Therefore Magnification (m) = -v/u= - 8.5 cm / -20 cm = +0.42

The positive value of image height indicates that the image formed is erect.

Therefore, the i mage formed is virtual, erect, and smaller in size.

Thus, Nature = Virtual and erect
Size = Smaller in size or diminished by the factor of 0.42
Position = 8.5 cm at the back of the mirror.

Regards