Question

an object is kept at a distance of 5 cm in front of the convex mirror of focal length 10 cm calculate the position and nature of the image formed

Answer 1

Object distance U = -5 cm(to the left of mirror)

Image distance V = ?

Focal length f = 10 cm(convex mirror)

By using the mirror formula we have

1/V + 1 / U = 1/f

1/V + 1/-5 = 1/10

1 / V = 1/10 + 1/5

1/V = 3/10 or V = 10/3

V = +3.33cm

Position of the image is at a distance of 3.33 cm from the convex mirror on its right side(behind the mirror). Since the image is formed behind the convex mirror, therefore the nature of image is virtual and erect.

Magnification m = - V/U

So m = - (3.33)/-5

m = +0.66

Answer 2

Explanation:

u=-5 cm, f=10 cm

We know that

v

1

+

u

1

=

f

1

⇒

v

1

+

(−5)

1

=

10

1

⇒

v

1

=

20

1

+

5

1

=

10

3

∴v=

3

10

cm=3.33cm

The position of the image is 3.33 cm behind the convex mirror.

Magnification, m=−

u

v

=−

−5

3.33

=0.66

The image is virtual and erect.

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