A convex mirror of radius of curvature 1.6 m has an object placed at a distance of 1.0 m from it. Find the position of image and the magnification.​

The image is formed at 0.44 m in front of the mirror

magnification (M) = 0.44

Explanation:

Applying, mirror formular

1/f = (1/u)+(1/v) Equation 1

Where f = Focal length of the convex mirror, u = object distance, v = image distance.

Using the real is positive convection,

From the question,

Given: f = -1.6/2 = -0.8 m( The focal length of a convex mirror is vitual), u = 1.0 m

Substitute these values into equation 1

-1/0.8 = (1/v)+(1/1)

Solve for v

1/v = 1.25+1

1/v = 2.25

v = 1/2.25

v = 0.44 m

Hence the image is formed at 0.44 m in front of the mirror

Magnification (M) = v/u

m = 0.44/1

m = 0.44

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.