Question

A convex mirror produces a magnification of 1/2 when the object is placed at a distance of 60 cm from it . where should the object be place so that size of the image becomes 1/3 of the object​

Answer 1

Answer:

The object should be placed in front of the mirror, at a distance of 120 cm from it.

Explanation:

m = + 1/2

u = -60 cm

So,

-v/u = + 1/2

or v = -u/2 = -(-60)/2 = 30 cm

1/f = 1/v + 1/u

1/f = 1/30 - 1/60 = (2-1)/(60) = 1/60

Therefore, f = + 60 cm

Now, m = + 1/3

-v/u = 1/3

-v = u/3

v = -u/3

1/f = 1/v + 1/u

1/60 = 1/-u/3 + 1/u = -3/u + 1/u = (-3+1)/u = -2/u

u = -2*60 = -120 cm

u = -120 cm

Therefore, the object should be placed in front of the mirror, at a distance of 120 cm from it.

Answer 2

Given:

• Convex mirror with magnification(m) of 1/2, i.e. 0.5.
• Object Distance(u) = 60cm.

Find:

• Object distance(u) where image size becomes 1/3 of the object.

Solution:

u = -60cm.

m = -v/u = +0.5 (Magnification of a convex mirror is always positive).

-v/-60 = 0.5.

v = 0.5*60.

v = 30 cm.

According to Mirror Formula:

1/f = 1/v + 1/u.

1/f = 1/30 + 1/60.

1/f = 2/60 - 1/60.

1/f = 1/60.

f =  60 cm.

Focal length(f) = 60 cm.

Let h be the size of the object.

Size of the new image(h') is h/3.

m = h'/h.

m = h/3 * 1/h.

m = +1/3.

m = -v/u.

+1/3 = -v/u.

-3v = u.

According to Mirror Formula:

1/f = 1/v + 1/u.

1/60 = 1/v + 1/-3v.

1/60 = 1/v - 1/3v.

1/60 = 3/3v - 1/3v.

1/60 = 2/3v.

3v = 2*60.

3v = 120.

v = 120/3 cm = +40 cm

Object Distance(u) = 3*40 = 120 cm.