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At what distance from a cancave mirror of focal length 10cm should an object 2cm long be placed in order to get an erect image 6cm tall?.....
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Answered by
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concave mirror , f is negative , f = -10 cm
u = ?
h = 2 cm
erect image so h' is positive, h' = 6 cm
so m = h'/h = 3
as m = -v/u = 3 => v = -3 u
mirror equation is 1/v + 1/u = 1/f
1/(-3u) + 1/u = 1/(-10)
2/(3u) = -1/10
u = -20 /3 cm = -6.67 cm and v = -3 u = 20 cm
v is positive, so image is behind the mirror and is virtual.
in concave mirror , an erect image is obtained when the object is between the pole and the focus. it is magnified and is virtual. so you can verify from values of u, v and f.
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u = ?
h = 2 cm
erect image so h' is positive, h' = 6 cm
so m = h'/h = 3
as m = -v/u = 3 => v = -3 u
mirror equation is 1/v + 1/u = 1/f
1/(-3u) + 1/u = 1/(-10)
2/(3u) = -1/10
u = -20 /3 cm = -6.67 cm and v = -3 u = 20 cm
v is positive, so image is behind the mirror and is virtual.
in concave mirror , an erect image is obtained when the object is between the pole and the focus. it is magnified and is virtual. so you can verify from values of u, v and f.
hope it helps you. . . . . follow me. . . . . mark as a brainlist. . . . .
Answered by
9
Hope this help........
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