(i)A pin of length 2 cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1 cm is formed at a distance of 40 cm from the pin then focal length of lens and its distance from pin are what
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h = 2 cm h' = 1 cm
image is inverted => image is real.
m = - v/u = h'/h = -1/2
u = - 2 v
u is negative and v is positive
-u + v = 40 cm
3 v = 40 cm => v = 40/3 cm
u = - 80/3 cm
1/f = 1/v - 1/u = 3/40 + 3/80 = 9/80
f = 80/9 cm
image is inverted => image is real.
m = - v/u = h'/h = -1/2
u = - 2 v
u is negative and v is positive
-u + v = 40 cm
3 v = 40 cm => v = 40/3 cm
u = - 80/3 cm
1/f = 1/v - 1/u = 3/40 + 3/80 = 9/80
f = 80/9 cm
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Answered by
17
h = 2 cm h' = 1 cm
m = - v/u = h'/h = -1/2
u = - 2 v
-u + v = 40 cm
2v + v = 40 cm
3 v = 40 cm
v = 40/3 cm
u = - 80/3 cm
1/f = 1/v - 1/u = 3/40 + 3/80 = 9/80
f = 80/9 cm
m = - v/u = h'/h = -1/2
u = - 2 v
-u + v = 40 cm
2v + v = 40 cm
3 v = 40 cm
v = 40/3 cm
u = - 80/3 cm
1/f = 1/v - 1/u = 3/40 + 3/80 = 9/80
f = 80/9 cm
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