Question

An object 1 cm high is place on the axis and 15 cm from a concave mirror of focal length 10 cm. Find the position nature magnification and size of the image draw the ray diagram

Answer 1

Hii

Here is your answer -

u=15cm

f=10cm

v=?

so

1/f=1/v+1/u

1/10=1/v+1/15

1/10— 1/15=1/v

3-2\30= 1/v

1/30 = 1/v

30 =v

so the image will be larger in size real & inverted & image is formed beyond c

&

magnification (m)= –v/u

= –30/15

=–2

so m=h'/h

–2= h'/1

h'=–2

Answer 2

Solution

u=−15cm

f=−10cm

h=1cm

From mirror formula

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

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

$\frac{1}{v}= - \frac{1}{10} -(- \frac{1}{15} )$

$v = -30 cm$

$\Rightarrow$$30 cm$ is front of mirror

The image is formed at a distance 30cm on the left side of the mirror from the pole.

$m = \frac{h'}{h} = \frac{-v}{u}$

$h' = \frac{-v}{u} * h$

$h' = -\frac{(-30)}{-15} *1$

$h' = -2 cm$

The image size is 2cm and it is real , magnified and inverted.

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