Question

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

Answer 1

The focal length of the concave mirror can be written as

1/f = 1/v + 1/u

Where

f = focal length of the mirror

v = distance of the image from the pole

u = Distance of the object from the pole

Given: The image is formed in front of the mirror, so the image is real and is at a distance of 30 cm from the pole.

∴ v = -30cm

Magnifying factor (m) = m/u = -y/x

y = Image height

x = Object’s height

m = -5/10

Since m = v/u

-5/10 = v/u

-5/10 = (-30)/u

∴ u = 60cm

Since object distance is always negative, so u = -60cm

Apply mirror formula as

(1/ v) + (1/u) = 1/f

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

-1/60 – 1/30 = 1/f

-90/1800 = 1/f

1 /(- 20) = 1/f or

f = – 20 cm

The focal length lies in front of the mirror. Therefore, the sign is negative.

Answer 2

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The focal length of the concave mirror can be written as

1/f = 1/v + 1/u

Where

f = focal length of the mirror

v = distance of the image from the pole

u = Distance of the object from the pole

Given: The image is formed in front of the mirror, so the image is real and is at a distance of 30 cm from the pole.

∴ v = -30cm

Magnifying factor (m) = m/u = -y/x

y = Image height

x = Object’s height

m = -5/10

Since m = v/u

-5/10 = v/u

-5/10 = (-30)/u

∴ u = 60cm

Since object distance is always negative, so u = -60cm

Apply mirror formula as

(1/ v) + (1/u) = 1/f

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

-1/60 – 1/30 = 1/f

-90/1800 = 1/f

1 /(- 20) = 1/f or

f = – 20 cm

The focal length lies in front of the mirror. Therefore, the sign is negative.

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