A student focus the image of a candle fame on a white screen using a convex lens. He noted down the position of the candle, screen and the lens as under:
Position of candle= 12cm
Position of convex lens= 50cm
Position of screen= 88cm
a) What is the length of the focal length?
b) Where will the image be formed if he shifts the candle towards the leans at a position of 31cm.
c) What will be the nature of the image formed, if he further shifts the candle towards the lens.
d)Draw a ray diagram to show the formation of the image in case (c) as said above...
PrincessManshi:
sorryy.. ths cums in physics
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a)
u = -(50 - 12) = - 38 cm
v = 88-50 = 38 cm
1/v - 1/u = 1/f
1/38 - 1/-38 = 1/f
f = 19 cm
b)
u = -(50-31) = -19 cm v = ? f = 19 cm
1/v = 1/u+1/f = 0 => Image will form at infinity, as the candle is at the focus.
c)
Let us say u = - 3f/4
1/v = 1/u + 1/f = -4/3f + 1/f = -1/3f
v = - 3f => a virtual magnified erect image is formed on the same side as the candle.
d) see diagram.
u = -(50 - 12) = - 38 cm
v = 88-50 = 38 cm
1/v - 1/u = 1/f
1/38 - 1/-38 = 1/f
f = 19 cm
b)
u = -(50-31) = -19 cm v = ? f = 19 cm
1/v = 1/u+1/f = 0 => Image will form at infinity, as the candle is at the focus.
c)
Let us say u = - 3f/4
1/v = 1/u + 1/f = -4/3f + 1/f = -1/3f
v = - 3f => a virtual magnified erect image is formed on the same side as the candle.
d) see diagram.
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sir .. what is -(50 - 12)
what is AB in the given diagram..???
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