Question

An object is placed 60 cm in front of a convex mirror. The virtual image formed by the mirror is located at 30 cm behind the mirror. What is the object's magnification??​

Answer 1

Answer:

object distance(u)=60cm

image distance(v)=30cm

by sign convention

u=-60cm

v=30cm

magnification=-v/u

substituting values in formula

we get,

-(-60)/30

60/30=2

magnification =2

therefore image is erect,virtual.

Answer 2

Given: An object is placed 60 cm in front of a convex mirror. The virtual image formed by the mirror is located at 30 cm behind the mirror.

To find: The object's magnification.

Solution:

• The linear magnification produced by a spherical lens is the ratio of the size of the image formed by the lens to the size of the object, both measured perpendicular to the principal axis.
• The distances above the principal axis are taken as positive and below as negative. Virtual images are erect and hence, the linear magnification is positive.
• It is given by the formula,

        $m = \frac{v}{u}$

• Here, m is the magnification produced, v is the distance of the image and u is the distance of the object from the lens.
• On replacing the terms with the values given in the question,

        $m = \frac{30 cm}{60 cm}$

             $= \frac{1}{2}$

Therefore, the object's magnification is 1/2 or 0.5.