Question

A concave mirror produces a real image of height 2cm of an object of height 0.5cm place 10cm away from the mirror.Find the position of the image and focal length of the mirror?

Answer 1

here here is your answer mate...

although the magnification is...

hi/ho

then.... 2/0.5 = 4

then..

m = -v/u = 4 = -v / 10

then..

v = -40 cm

then the focal length will be...

1/f = 1/v + 1/u = 1/-40 + 1/10

then.. 1/f = 4-1/40 = 3/40

then..

f = 40/3 = 13.3333333

HOPe it helps you...

Answer 2

Answer:

Given:-

Distance of the object, (u) = -10 cm

Size of the Image, (I) = -2 cm

Size of the Object, (o) = 0.5 cm

To Calculate:-

(a) Distance of the image (v) = ???

(b) Focal Length (f) = ???

Solution:-

Formula to be used

1.   m = $\frac{I}{O}$

2.  m = - $\frac{v}{u}$

3.  $\frac{1}{v} +\frac{1}{v} =\frac{1}{f}$

Since, we are given I and O, we will first find out m and then from formula (2) we will find out v.

From the first formula,

⇒ $m=\frac{I}{O}$

⇒ $m=\frac{-2}{0.5}$

$=-4$

Putting this value of m in formula (2.), we get

⇒ $m=-\frac{v}{u}$

⇒ $-4=-\frac{v}{-10}$

$v=-40 \: cm$

Now, to find f we will use formula (3.)

⇒ $\frac{1}{v} +\frac{1}{v} =\frac{1}{f}$

⇒ $\frac{1}{f} =-\frac{1}{40} +(-\frac{1}{10})$

⇒ $\frac{1}{f} =-\frac{1}{40}-\frac{1}{10}$

⇒ $\frac{1}{f}=\frac{-1-4}{40}$

⇒ $\frac{1}{f}=\frac{-5}{40}$

$=-8 \: cm$

Distance of the image (v) = -40cm

Focal Length (f) = -8 cm