Question

A 4.5 c.m. needle is placed 12c.m. away from a convex mirror of focal length 15c.m. Give the location of the image and magnification

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Answer 1

Answer:

h=4.5

u= -12

f= 15

Explanation:

1/f=1/v+1/u

1/15=1/v+1/(-12)

1/v=1/15+1/12

1/v=12+15/15×12

1/v=27/15×12

1/v=9/5×12

1/v=3/5×4

1/v=3/20

v=20/3

m= -v/u

m= (-20/3)/-12

m=20/36

m=5/9

m=0.55

Answer 2

Given :-

$u=-12cm\\H=4.5\\f=-15cm$

What we need to find :-

$v=?\\H'=?\\m=?$

Procedure :-

Let us use the mirror formula.

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

Now let us plug in the values given.

$(\frac{1}{-15} +\frac{1}{-12} )=\frac{1}{v}\\\frac{1}{-3} (\frac{1}{5} +\frac{1}{4} )=\frac{1}{v}\\\frac{1}{-3} (\frac{4+5}{20} )=\frac{1}{v}\\\frac{1}{-3} (\frac{9}{20} )=\frac{1}{v}\\-(\frac{3}{20} )=\frac{1}{v}\\\\v=\frac{20}{3}cm$

Now let us obtain the magnitude.

$m=\frac{H'}{H} =-\frac{v}{u} \\m=\frac{H'}{4.5} =-(\frac{\frac{20}{3} }{-12} )\\\frac{H'}{4.5} =-(\frac{\frac{5}{3} }{-3} )\\\frac{H'}{4.5} =(\frac{1 }{9} )\\H' =(\frac{4.5 }{9} )\\H=0.5$

Now we need magnification.

$m=\frac{H'}{H}\\\\m=\frac{\frac{1}{2} }{\frac{9}{2} }\\\\m={\frac{1}{9} }} }$