Answers

2014-10-24T15:53:50+05:30
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
0
can i know which one is correct for sure
f = - 9.6 cm
not -6.4cm??
anywayz thakz a lot ankit
no no... i have corrected my mistake. please recheck my answer.... and welcome
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2014-10-24T16:23:51+05:30

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Height of the object, h1 = 2 cm
height of the image, h2 = -3 cm
object distance, u = -16cm
image distance = v

magnification =  \frac{h2}{h1} = \frac{-v}{u}
 
 \frac{-3}{2} = \frac{-v}{-16}

(-16)*(-3)=2*(-v)
v = - \frac{48}{2} =-24cm

From mirror formula 
 \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

So focal length is 9.6cm and image distance is 24cm(to the left of mirror)
1 5 1
i liked ur way of answering
thanks dude!!
yesss, i appreciate your way of answering....! :)
i appreciate u both cause u btoh are saving me from my physics teacher