Question

A concave mirror produces three times enlarged virtual image of an object placed at 10 cm in front of it. Where is the image located?​

Answer 1

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m = 3 (virtual image)

m = 3 (virtual image)u = - 10 cm

m = 3 (virtual image)u = - 10 cmR = ?

m = 3 (virtual image)u = - 10 cmR = ?We know that

m = 3 (virtual image)u = - 10 cmR = ?We know thatm = - (v/u)

m = 3 (virtual image)u = - 10 cmR = ?We know thatm = - (v/u)3 = -v/(-10)

m = 3 (virtual image)u = - 10 cmR = ?We know thatm = - (v/u)3 = -v/(-10)V = 30 cm

m = 3 (virtual image)u = - 10 cmR = ?We know thatm = - (v/u)3 = -v/(-10)V = 30 cmAnd

m = 3 (virtual image)u = - 10 cmR = ?We know thatm = - (v/u)3 = -v/(-10)V = 30 cmAnd1/v + 1/u = 1/f

m = 3 (virtual image)u = - 10 cmR = ?We know thatm = - (v/u)3 = -v/(-10)V = 30 cmAnd1/v + 1/u = 1/f1/30 + 1/(-10) = 1/f

m = 3 (virtual image)u = - 10 cmR = ?We know thatm = - (v/u)3 = -v/(-10)V = 30 cmAnd1/v + 1/u = 1/f1/30 + 1/(-10) = 1/f-20/300 = 1/f

m = 3 (virtual image)u = - 10 cmR = ?We know thatm = - (v/u)3 = -v/(-10)V = 30 cmAnd1/v + 1/u = 1/f1/30 + 1/(-10) = 1/f-20/300 = 1/ff = - 15 cm

m = 3 (virtual image)u = - 10 cmR = ?We know thatm = - (v/u)3 = -v/(-10)V = 30 cmAnd1/v + 1/u = 1/f1/30 + 1/(-10) = 1/f-20/300 = 1/ff = - 15 cmRadius of curvature = R = 2f

m = 3 (virtual image)u = - 10 cmR = ?We know thatm = - (v/u)3 = -v/(-10)V = 30 cmAnd1/v + 1/u = 1/f1/30 + 1/(-10) = 1/f-20/300 = 1/ff = - 15 cmRadius of curvature = R = 2f= 2 × (-15) = - 30 cm

Explanation:

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Answer 2

➡Given,

Magnification, m=3

Object distance, u=10cm

We know,

Magnification, m=−uv

➡3= −10−v

⇒ v=30cm

From mirror formula,

f1 = v1 + u1

f1 = 301 − 101

➡= 30×10

➡10−30

➡= 300−20

f=−15cm

Focal length, f=−15cm

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