Plz solve this que??????
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check this answer and tell me
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tanisha50:
John object ki position kya h
-30cm
Answered by
1
height of object = 1.2 m
focal length, f = -20 cm
image distance , v = -60 cm
mirror formula
1/v + 1/u = 1/f
1/u = 1/f - 1/v
1/u = 1(-20) - 1/(-60)
1/u = 1/-20 + 1/60
1/u = -1/20 + 1/60
1/u = (-3 +1) / 60
1/u = -2/60
u = -60/2
u = -30 cm
as we know
m = height of image / height of object .................(i)
and also
m = -v/u............(ii)
from (i) and (ii) we get
hi/ho = -v/u
hi/ 1.2 = -( -60/-30)
hi/ 1.2 = -2
hi = -2.4 m
position of the object =
as the focal length is 20 and object distance from the mirror is 30 cm
so object distance is more than focal length
hence object is in between the center of curvature and focus
focal length, f = -20 cm
image distance , v = -60 cm
mirror formula
1/v + 1/u = 1/f
1/u = 1/f - 1/v
1/u = 1(-20) - 1/(-60)
1/u = 1/-20 + 1/60
1/u = -1/20 + 1/60
1/u = (-3 +1) / 60
1/u = -2/60
u = -60/2
u = -30 cm
as we know
m = height of image / height of object .................(i)
and also
m = -v/u............(ii)
from (i) and (ii) we get
hi/ho = -v/u
hi/ 1.2 = -( -60/-30)
hi/ 1.2 = -2
hi = -2.4 m
position of the object =
as the focal length is 20 and object distance from the mirror is 30 cm
so object distance is more than focal length
hence object is in between the center of curvature and focus
position of the object dear???
plz can u tell
yes
so what's it??
i edited my answer u can see in it
its -30 i think
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