Question

An object 2 cm high is placed at a distance of 16 cm from a concave mirror which produces a 5]

real image 3 cm high.

(iFind the position of the image.

i)What is the focal length of mirror?​

Answer 1

Explanation:

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Answer 2

GIVEN :-

$\textsf{ •Object height ,h=+2 </u></strong><strong><u>cm}$

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

$\textsf{•Object distance,u=-16cm }$

TO FIND OUT:-

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

$</u><u>\</u><u>purple</u><u>{</u><u> \textsf{•Focal length of the image (f)=?}</u><u>}</u><u>$

SOLUTION:-

$\green{\textsf{Finding Position of the image (v)}}$

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

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

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

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

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

•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)

$\purple{ \textsf{Finding Focal length of the image (f)}}$

$\textsf{•Using mirror formula,}$

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

$\textsf{•Putting the value ,we get}$

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

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

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