Question

An object of size 3 cm is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm . Find the position and size , also nature of image.​

Answer 1

Answer:

Given: u=−20 cm

           f=R/2=15 cm

           Size of image, h=5 cm

Let size of image be h  

From mirror formula,  

1/u + 1/v = 1/f

1/-20 + 1/v = 1/15

v = 60/7 = 8.57 cm

The image is formed 8.57 cm behind the mirror. It is virtual and erect.

Magnification, m = −v/u = 8.57/20 = 0.428

Also, m = h'/h = h'/5

h'/5 = 0.428

h' = 2.14 cm

Answer 2

Answer:

• u = - 20 cm
• radius = 30 cm
• Size of object = h = 3 cm

Radius of curvature = 30 cm

Hence, focal length = 15 cm

Mirror Formula = 1/f = 1/u + 1/v

Hence, 1/15 = 1/-20 + 1/v

Solving, 1/v = 1/15 + 1/20

1/v = (15 + 20)/300

1/v = 35/300

1/v = 7/60

v = 60/7 = 8.6 cm — [Position]

Now, h(image)/h(object) = -v/u

h(image)/3 = -8.6/-20

h(image) = 8.6/20 * 3

h(image) = 86/200 * 3

h(image) = 43/100 * 3

h(image) = 123/100 = 1.23 cm — [Size]

As, the image image distance is positive, it indicates that the images formed behind the mirror, and is virtual and erect. — [Nature of image]