Question

An object l0cm in length is placed at a distance of 30cm
in front of a convex mirror of radius of curvature 50cm
Find the position of the image, its nature and size​

Answer 1

Given

• Height of an object (ho) = + 10cm
• Object distance (u) = - 30cm
• Radius of curvature (R) = 50cm

Find out

• Position , nature and size of image

Solution

➟ Focal length = Radius of curvature/2

➟ f = R/2

➟ f = 50/2 = + 25 cm

Apply Mirror Formula

➟ 1/u + 1/v = 1/f

➟ - 1/30 + 1/v = 1/25

➟ 1/v = 1/25 + 1/30

➟ 1/v = 6 + 5/150

➟ 1/v = 11/150

➟ v = 150/11 = + 13.6 cm

Hence, image distance is + 13.6 cm

Magnification : It is the ratio of height or size of image to the height or size of object.

➟ m = -v/u = hi/ho

➟ -v/u = hi/ho

➟ - 13.6/(- 30) = hi/10

➟ 13.6 × 10 = 30 × hi

➟ 136 = 30hi

➟ hi = 136/30 = +4.5cm

Hence, size of image is + 4.5cm

• Nature of image
• Nature = Virtual and erect

Answer 2

Solution

f = R/2

f = 50/2 = + 25 cm

Apply Mirror Formula

1/u + 1/v = 1/f

- 1/30 + 1/v = 1/25

1/v = 1/25 + 1/30

1/v = 6 + 5/150

1/v = 11/150

v = 150/11 = + 13.6 cm

m = -v/u = hi/ho

-v/u = hi/ho

- 13.6/(- 30) = hi/10

13.6 × 10 = 30 × hi

136 = 30hi

hi = 136/30 = +4.5cm

Hence, size of image is + 4.5cm

Nature of image

Nature = Virtual and erect