An object 4cm high is placed at a distance of 6 cm in front of a concave mirror of focal length 12 cm. Find the position, nature and size of the image formed.
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we know mirror formula
(1/v)+(1/u)=(1/f)
1. (1/v)=(1/f)-(1/u)
(1/v)=(1/12)-(1/6)
= -(1/12)
So, v = -12cm that means the image will form 12cm away and behind the mirror.
2. image will be of imaginary nature as v is negative
3. By using magnification equation
M=(hi/ho)= -(v/u)
(hi/4)= -(-12/6)
hi=8cm
So the height of the image will be 8cm
(1/v)+(1/u)=(1/f)
1. (1/v)=(1/f)-(1/u)
(1/v)=(1/12)-(1/6)
= -(1/12)
So, v = -12cm that means the image will form 12cm away and behind the mirror.
2. image will be of imaginary nature as v is negative
3. By using magnification equation
M=(hi/ho)= -(v/u)
(hi/4)= -(-12/6)
hi=8cm
So the height of the image will be 8cm
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