Question

A 10 cm long stick is kept in front of a concave mirror having a focal length of 10 cm in such a way that the end of the stick closest to the pole is at a distance of 20 cm. What will be the length of the image?

Answer 1

Given: Focal length (f) = −10 cm    ...(Negative for concave mirror)

Near end distance (u₁) = −20 cm

Far end distance (u₂) = −(20 + 10) = −30 cm

Use the mirror formula:

`1/f = 1/v_1 + 1/u_1`

Step 1: Image of the nearest end (u₁ = 20 cm)

`1/-10 = 1/v_1 + 1/-20`

`1/v_1 =1/-10 - 1/-20`

`1/v_1 = -1/10 + 1/20`

`1/v_1 = -1/20`

v₁ ​= −20 cm

Step 2: Image of the farthest end (u₂ = −30 cm)

Use the mirror formula:

`1/f = 1/v_2 + 1/u_2`

`1/-10 = 1/v_2 + 1/-30`

`1/v_2 = -1/10 + 1/30`

`1/v_2 = (-3 + 1)/30`

`1/v_2 = -2/30`

`1/v_2 = -1/15`

v_2​ = −15 cm

Step 3: Length of the image:

The two image points form at −15 cm and −20 cm from the mirror.

Image length = |v₁ − v₂|

= ∣(−20) − (−15)∣

= |−5|

= 5 cm

The length of the image is 5 cm.