Question

qu : 1 An object is kept at a distance of 5 cm in front of convex mirror of focal length 10 cm. calculate the position and magnification of the image and state its nature.

qu: 2 An object is placed at a distance of 6 cms from a convex mirror of focal length 12 cm. find the position and nature of the image.

Answer 1

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 = 10/3V = +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
2)Object distance U = -6 cm(to the left of mirror)Image distance V = ?
Focal length f = 12 cm(convex mirror)
By using the mirror formula we have
1/V + 1 / U = 1/f
1/V + 1/-6 = 1/12
1 / V = 1/12 + 1/6
1/V = 3/12 or V = 12/3V = +4cmPosition of the image is at a distance of 4 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 = - (4)/-6m = +0.66

Answer 2

Answer:

For question 1

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

The image is virtual and erect.