Question

A 10 mm long pin is placed vertically in front of concave mirror. A 5 mm of long image of pin is formed at 30 cm in front of the mirror. The focal length of the mirror is​

Answer 1

Given:-

• Height of object ,ho = 10mm = 1 cm

• Image Size ,hi = 5mm = 0.5 cm

• Image distance, v = -30cm

To Find:-

• Focal length ,f

Solution:-

Using magnification Formula

• m = hi/ho = -v/u

Substitute the value we get

→ 0.5/1 = -30/u

→ u = -30×1/0.5

→ u = -30/0.5

→ u = -60 cm

the Object distance is -60 cm

Now, Using mirror Formula

• 1/v + 1/u = 1/f

Substitute the value we get

→ -1/30 + (-1/60) = 1/f

→ -1/30 -1/60 = 1/f

→ 1/f = -2-1/60

→ 1/f = -3/60

→ 1/f = -1/20

→ f = -20 cm

Therefore, the focal length of the mirror is -20cm.

Answer 2

Given:

• Size of pin, h = 10 mm = 1 cm
• Size of image formed of pin, h' = 5 mm = 0.5 cm
• Image distance from the mirror of pin, in front of mirror, v = - 30 cm

To find:

• Focal length of the mirror?

Solution:

☯ Let f be the focal length of mirror.

We know that,

h'/h = - v/u [ from formula to find magnification ]

⇒ u = - vh/h'

⇒ u = (- 30 × 1)/0.5

⇒ u = (- 30 × 10)/ 5

⇒ u = - 300/5

⇒ u = - 60 cm

∴ Object distance, u is - 60 cm

★ Now, Using mirror formula,

1/f = 1/v + 1/u

1/f = 1/(-30) + 1/(-60)

1/f = 1/20

f = - 20 cm

∴ Hence, Focal length of mirror is 20 cm.