Physics, asked by sabeelasiddiquie, 1 year ago

A CONCAVE MIRROR PRODUCE a real image of half the size of an object placed at 60 cm in front of it

Answers

Answered by Himankp
6
Initial obj. dist = -60(real object)
". magnification (m) = -1/2 (as both obj and image are real)
We know , m= -f/(u-f) [f=focal length]
f= -20cm
Now , obj dist = u'
magnification(m')=2 (as image is virtual of real object)
f=-20 cm
We know , m= -f/(u'-f)
2=20/(u'+20)
u'=-10cm
Hence , the obj should at a distance of 10 cm in front of the mirror.
Answered by ashishjha246
1

Answer:

magnification can also be written as m = f / f - u

Real image in case of concave mirror will have m = -1/2 (inverted image for erect object)

Plugging m = -1/2 and u = 60 cm we have -1/2 = f / f-u

So f = 20 cm

Now we expect m = 2 (virtual is also erect)

Again 2 = f/ (f-u)

Plugging 2 = 20 / (20 - u)

==> solving we get u = 10 cm

Object has to be placed with focal length at a distance 10 cm infront of the mirror.

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