Question

Find the size, Nature and position of images when

a. m=-1 d.m>-1

b.m<+1 e.m>+1

c.m<-1

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Answer 1

u=−15cm

u=−15cmf=−10cm

u=−15cmf=−10cmh=1cm

u=−15cmf=−10cmh=1cmFrom mirror formula

u=−15cmf=−10cmh=1cmFrom mirror formulaf

u=−15cmf=−10cmh=1cmFrom mirror formulaf1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 =

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 +

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 =

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 −

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 =

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −10

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101 −

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101 − −15

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101 − −151

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101 − −151

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101 − −151

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101 − −151 v=−30cm

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101 − −151 v=−30cmThe image is formed at a distance 30cm on the left side of the mirror from the pole.

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101 − −151 v=−30cmThe image is formed at a distance 30cm on the left side of the mirror from the pole.

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101 − −151 v=−30cmThe image is formed at a distance 30cm on the left side of the mirror from the pole.

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101 − −151 v=−30cmThe image is formed at a distance 30cm on the left side of the mirror from the pole. h=−2cm

u=−15cmf=−10cmh=1cmFrom mirror formulaf1 = u1 + v1 v1 = f1 − u1 v1 = −101 − −151 v=−30cmThe image is formed at a distance 30cm on the left side of the mirror from the pole. h=−2cmThe image size is 2cm and it is real and inverted.

Explanation:

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