Question

An image 5cm in height is placed at a distance of 15cm in front of a convex mirror of radius of curvature of 40cm.Find the position of the image and nature of the image.​

Answer 1

Answer:

Radius of curvature (R) = 30 cm

f = R/2 = 30/2 = 15 cm

u = -20 cm,

h= 5 cm.

1/v +1/u = 1/f

1/v = 1/f – 1/u

1/v = 1/15 – 1/-20

v = 8. 57 cm

Image is virtual and erect and formed behind the mirror.

m = -v/u

m = hi/ho

hi/5= 8.57/20

= 0.428

hi = 0.428 x 5 = 2.14cm

Position of Image: Behind the mirror.

Nature of Image: Virtual and Erect.

Size of the Image: Diminished.

Answer 2

Answer:

Step-by-step explanation:

Radius of curvature (R) = 30 cm

f = R/2 = 30/2 = 15 cm

u = -20 cm,

h= 5 cm.

1/v +1/u = 1/f

1/v = 1/f – 1/u

1/v = 1/15 – 1/-20

v = 8. 57 cm

Image is virtual and erect and formed behind the mirror.

m = -v/u

m = hi/ho

hi/5= 8.57/20

= 0.428

hi = 0.428 x 5 = 2.14cm

Position of Image: Behind the mirror.

Nature of Image: Virtual and Erect.

Size of the Image: Diminished.