Finding the image distance and magnification of an object infront of a Convex mirror?

Question

A 6.50 cm pencil is placed upright at a distance 54.0 cm away from the center of a convex mirror. The mirror's radius of curvature is 45.0 cm. Calculate the image distance.
What is the magnification?

Answer 1

Radius of curvature is 45 cm, therefore the focal length is -22.5 cm. (negative because the mirror is convex).

u = image distance = 54 cm.

1/u + 1/v = 1/f

Putting in the figures and solving gives v = -15.9 cm

Hence the image is virtual.

m = -v/u

m = -(-15.9)/54 = 0.3

the magnification, m, is less than unity, hence the image is diminished. The sign for m is positive, hence the image is erect.

Answer 2

Image Distance Formula

Answer 3

lets p the distance from the object to the summit of mirror.
lets p' the distance from the image to the summit of mirror.
the focal distance is f = radius/2 = -0.45/2= -0.225m (negative
since the mirror is convex)

you use the formula

1/p'+1/p =1/f here f negative -0.225m
p is positive = 0.54

So 1/p' = -1/0.225-1/0.54= -4.444-1.85=-6.3
and p' = -0.159m from the summit

The magnification is -0.159/-0.54= 0.294

image virtual and in the same direction

Answer 4

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In S.I. system , the focal length of convex mirror is positive.

Focal length =(1/2)radius of curvature=(1/2)45 =22.5 cm

Assuming the object is on left hand side and mirror is on right hand side.

The object distance 'u'= - 54 cm

Using mirror formula, 1/v + 1/u = 1/f

1/v + 1/ (-54) =1/22.5

1/v =1/22.5 +1/54=153 / (45*54)

Image distance v = 15.882 cm

Positive image distance means image is behind the mirror on the right of mirror,between pole and focus of convex mirror.
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In S.I. system, magnification m= - (v/u) = -(- 45 /153) = 0.294

Magnification 'm' =0.294

' m' is positive and less than 1

Positive magnification means image is upright, virtual, smaller in size(diminished),cannot be obtained on a screen

size of image= magnification 'm' * size of object

size of image =[45/153]*6.5 cm=1.911 cm

Image is 1.911 cm high, upright, behind the mirror
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Answer 5

1/u + 1/v = 1/f gives 1/u -- 1/v = --1/f also object : image = n : 1 giving u/v = n or v = u/n then 1/u -- n/u = -1/f or (n -- 1)/u = 1/f whence u = (n -- 1)f object distance is (n --1) times focal length f.