A 3 cm tall object is placed perpendicular to principal axis of a convex of a Focal length 15cm the distance of an object from the axis is 18 cm find the mature, position and size and magnification of image
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hi mate...
your answer is in the pic...
your answer is in the pic...
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gurdeenkaur31:
ya
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next time read the question properly bhai
I m so sorry actually it did not mentioned that if it's a mirror or a lense
no problem...
what is the answer
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that answer is correct
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Answered by
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height of object(h)=3 cm
focal length(f)=15cm
object distance (u)= -18cm
by mirror formula,
1/f=1/v+1/u
1/v=1/f-1/u
1/v=1/15-1/(-18)
1/v=1/15+1/18
=(6+5)/90
=11/90
v=90/11
v=8.18cm
the nature of the image formed is virtual,erect and diminished and smaller in size than object.
and it is 8.18cm away from the mirror.
height of image=h¹
h/h¹=-v/u
3/h¹=-8.18/(-18)
h¹=3(18)/8.18
h¹=(54)/8.18
h¹=-6.601cm
height of image formed is 6.601cm.
magnification (m)=h/h¹
=3/6.601
=0.45
hope it helps....
please mark as brainlist.....
focal length(f)=15cm
object distance (u)= -18cm
by mirror formula,
1/f=1/v+1/u
1/v=1/f-1/u
1/v=1/15-1/(-18)
1/v=1/15+1/18
=(6+5)/90
=11/90
v=90/11
v=8.18cm
the nature of the image formed is virtual,erect and diminished and smaller in size than object.
and it is 8.18cm away from the mirror.
height of image=h¹
h/h¹=-v/u
3/h¹=-8.18/(-18)
h¹=3(18)/8.18
h¹=(54)/8.18
h¹=-6.601cm
height of image formed is 6.601cm.
magnification (m)=h/h¹
=3/6.601
=0.45
hope it helps....
please mark as brainlist.....
ok
u can report my answer
formula is wrong
this is mirror formula but in numerical have to use lens formula
you don't write lens or mirror so I answer by taking mirror
that's your mistake
ok
sorry
now we need to delete the answer and correct it!!
no problem tanayajagadale
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