Question

A 5cm long needle lies along the principal axis of a concave mirror of focal length of 20 cm in such a way that the end closer to the pole is 40 cm away from it. Find the length of the image of the needle formed by the mirror.

Answer 1

$\blue{\textbf{Answer}}$:-

➮let AB is the given length of needle.

The end B if of the needle always lie at the $\blue{\texttt{centre of curvature C}}$ to the mirror as per given condition.

➮Therefore,

its image also $\blue{\texttt{formed at C}}$ .

Let A1 is the image of the end of A of needle .

➮For the $\green{\textbf{end A}}$ ,

U = -45 cm

f = - 20cm

v = ??

$\blue{\texttt{applying mirror formula}}$,

$\frac{1}{f} = \frac{1}{u} - \frac{1}{v}$

➮and subtituting the values ,

$\frac{1}{ - 20} = \frac{1}{ - 45} + \frac{1}{v}$

$\frac{1}{ v } = \frac{1}{ 45} - \frac{1}{20}$

= $\frac{4-9}{ 180}$

= $\frac{-5}{180}$

=$\frac{-1}{36}$

➮V = -36 , v=PA1

therefore, V = $\green{\textbf{36cm}}$

hence,

➮$\blue{\textbf{length of image }}$

BA1 = PB - PA1

= 40 cm - 36 cm

= 4cm

➮$\blue{\texttt{length of image }}$ of the needle formed by the mirror is 4cm.

Answer 2

Answer =

length=4cm.

Explaination ,

focal length = -20cm

u = -45

by mirror formula,

1/f = 1/u + 1 / v

= 1/-20 = 1/-45 +1/v

= 1/36 = 36cm

lets findout the length of image

= 40 - 36cm = 4cm