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an object 2cm high is placed at a distance of 16cm from a concave mirror which produces a real image 3cm high.      what is focal length of the mirror and position of the image??????

Answer :

ankitkumar0102
size of the object = h1 = 2 cm
size of the image = h2 = -3 cm
u = obj distance = -16 m

Therefore from formula -v/u = h2 / h1, we have
-v/(-16) = -3/2
v/16 = -3/2
v = -3* 16/2
v = - 24 cm

From mirror formula = 1/v + 1/u = 1/f
=> 1/-24 + 1/-16 = 1/f
=> 1/f = (-2 - 3)/ 48
=> 1/f = -5 / 48
=> f = - 48/5
= > f = - 9.6 cm
TPS
height of the object, h1 = 2 cm
height of the image, h2 = -3 cm
object distance, u = -16cm
image distance = v

magnification = [tex] \frac{h2}{h1} = \frac{-v}{u} [/tex]
 
⇒[tex] \frac{-3}{2} = \frac{-v}{-16} [/tex]

⇒[tex](-16)*(-3)=2*(-v)[/tex]
⇒[tex]v = - \frac{48}{2} =-24cm[/tex]

From mirror formula 
[tex] \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \\ \\ \frac{1}{-16} + \frac{1}{-24} = \frac{1}{f} \\ \\ \frac{3+2}{-48} = \frac{1}{f} \\ \\ \frac{5}{-48} = \frac{1}{f} \\ \\ f= \frac{-48}{5} = -9.6cm [/tex]

So focal length is 9.6cm and image distance is 24cm(to the left of mirror)

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