Question

an object 2cm high is placed at a distance of 16cm from a concave mirror, which produces 3cm high inverted image.what is the focal length of mirror?also find the position of image.
PLEASE ANSWER​

Answer 1

Answer:

Here h

1

= 2 cm, u = -16 cm, h

2

= -3 cm (Since image is real and inverted.)

∵m=

h

1

h

2

=−

u

v

∴v=

h

1

−h

2

u=

2

3

×(−16)=−24cm

f

1

=

v

1

+

u

1

=−

24

1

−

16

1

=

48

−2−3

=

48

−5

;f=

5

−48

=−9.6cm

Answer 2

GIVEN :-

$\textsf{ •Object height ,h=+2 cm}$

$\textsf{•Image height ,h'=-3 cm (real image hence inverted)}$

$\textsf{•Object distance,u=-16cm }$

TO FIND OUT:-

$\textsf{•Position of the image (v)=?}$

$\textsf{•Focal length of the image (f)=?}$

SOLUTION:-

i)POSITION OF IMAGE

$\textsf{•From the expression for magnification }$

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

$\implies \bf \: v = - u×\frac{h'}{h}$

$\implies \bf v=-(-16)× \frac{(3)}{2}$

$\bf \implies \: v = - 24 \: cm$

•The image is formed at distance of 24 cm in front of the mirror (negative sign means object and image are on the same side)

ii) FOCAL LENGTH OF THE MIRROR

$\textsf{•Using mirror formula,}$

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

$\textsf{•Putting the value ,we get}$

$\bf \implies \frac{1}{f} = \frac{ \: 1}{ - 16} + \frac{ \: 1}{ - 24}$

$\bf \implies \: f = - \frac{48 }{5} = - 9.6 \: cm$

$\bf \implies \frac{1}{f} = - \frac{3 + 2}{48} = - \frac{5}{48}$