Question

A 10 cm long stick is kept in front of a concave mirror having focal lengthof 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: 10 cm). Solve the example.

Answer 1

1/f=1/v+1/u F=-10; U=-20 
1/v=1/f-1/u
1/v=-1/10+1/20
1/v=-2+1/20
1/v=-1/20
v=-20cm

m=-v/u=h_/h
-20/20=h_/10
h_=10 
Hope u understand

Answer 2

Answer:

10 cm.

Explanation:

Sign Convention used:

All the distances which measured along the right of the mirror and above the principal axis of the mirror are taken as positive and all the distances which measured along the left of the mirror and below the principal axis of the mirror are taken as negative

Given:

• Length of the stick, $l_o = 10\ cm$
• Focal length of the concave mirror, $f=-10\ cm$.
• Distance of the end of the stick closest to the pole, $u=-20\ cm$.

Let $v$ be the distance of the image of the stick from the mirror and $l_i$ be the height of the image of the stick.

Using mirror equation,

$\dfrac 1f = \dfrac 1v +\dfrac 1u\\\dfrac 1v = \dfrac 1f -\dfrac 1u\\\dfrac 1v = \dfrac 1{-10} -\dfrac 1{-20}\\=\dfrac {-1}{10} +\dfrac 1{20}\\=\dfrac {-2+1}{20}\\=\dfrac{-1}{20}\\\Rightarrow v=-20\ cm.$

Now, the magnification of the mirror is given by

$m=\dfrac{-v}u = \dfrac{-(-20)}{-20}=-1.$

The magnification of the mirror is also defined in terms of the lengths(heights) of the stick and its image,

$m=\dfrac{l_i}{l_o} = \dfrac{l_i}{10}\\-1=\dfrac{l_i}{10}\\l_i=-10\ cm.$

Thus, the length of the image of the stick is 10 cm, the negative sign indicates that the image of the stick is inverted.