Answer the question 6 and 11
Answers
6) Magnigfication = hi / ho = -v / u
We can assume it to be virtual because in a concave mirror, it is enlarged only when it is virtual. Otherwise, if you refer to the ray diagrams, they are all diminished.
The remaining procedure is correct, wherein we get -v /u = 2
So, if v = - 2u.
Using the mirror equation,
1 / v + 1 / u = 1 /f
= -1 / 2u + 1 / u = 1 / -30
u = - 15 cms. which is between the focal length and the mirror. This proves that the image will be virtual and enlarged.
11) Given :
height of object=h0=5cm
focal length of lens=10cm
object distance=-30cm after sign conventions.
image distance=?
By lens formula :
1/f=1/v-1/u
1/v=1/f+1/u
=1/10-1/30
=3-1/30
2/30
v=30/2= 15cm
magnification=m=v/u
m=15/-30 =-1/2= -0.5cm
m=hi/ho
hi=mxho
=-0.5x 5
= -2.5cms
Thus object is placed beyond C and image formed is in between F and c.
Inverted, real and diminished
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