Question

An object 5 cm high forms a virtual image of 1.25 cm high, when placed in front of a convex mirror at a distance of 24 cm. Calculate: (a) the position of the image, (b) the focal length of the convex mirror.

Answer 1

Given :-

• object height (h) = 5cm
• image height (h') = 1.25cm
• object distance (u) = -24cm
• position of image (v) = ?
• focal length (f) = ?

(a) Magnification (M) = (h')/h = (-v)/u

Substituting the known values :-

1.25/5 = -v/(-25)

5(-v) = 1.25(-25)

-5v = -30

v = -30/-5

v = 30/5

v = 6cm

•°• Image is formed 6cm in front of the mirror.

(b) Mirror formula = 1/v + 1/u = 1/f

1/f = 1/6 + 1/(-24)

1/f = 1/6 - 1/24

1/f = (24 - 6)/144

1/f = 18/144

1/f = 9/72

1/f = 1/8

f = 8cm

•°• Focal length of mirror is 8cm

Answer 2

Given

● Height of object (h) = h_o = 5 cm.

● Height of image (h')= h_i = 1.25 cm.

● Distance of object (u) = -24 cm.[∵ Object is 24 cm in front of mirror]

To find

● Position of image (v)

● Focal length of convex mirror (f)

Solution

a) Using magnification equation :

★ m = hi/ho = -v/u

⇒ 1.25/5 = -v/-24

⇒ -5v = -30

⇒ v = -30/-5

⇒ v = 6 cm

∴ Image is 6 cm in front of the mirror.

___________________________

Again,

b) Using mirror formula :

★ 1/f = 1/v + 1/u

⇒ 1/f = 1/6 + 1/-24

⇒ 1/f = (-4 + 1)/-24

⇒ 1/f = -3/-24

⇒ -3f = -24

⇒ f = -24/-3

⇒ f = 8 cm

∴ Focal length of convex mirror = 8 cm.