An up-pointing arrow that is 1.0 cm tall is placed at a distance of 14.0 cm to t

Question

An up-pointing arrow that is 1.0 cm tall is placed at a distance of 14.0 cm to the left of a thin converging lens. The focal length of the lens is 6.0 cm. What is the orientation and height of the image of the arrow?

pointing up, height = 1.3 cm

pointing down, height = 0.50 cm

pointing down, height = 0.75 cm

pointing up, height = 0.50 cm

An up-pointing arrow that is 1.0 cm tall is placed at a distance of 14.0 cm to the left of a thin converging lens. The focal length of the lens is 6.0 cm. What is the orientation and height of the image of the arrow?

pointing up, height = 1.3 cm

pointing down, height = 0.50 cm

pointing down, height = 0.75 cm

pointing up, height = 0.50 cm

Explanation / Answer

for converging lens

focal length f is positive

then using lens equation

1/f = 1/do + 1/di

1/6 = 1/14 + 1/di

1/di = 1/6 - 1/14 =

di = 10.5 cm

magnification m = -di/do = hi/ho

image height hi = -di*ho/do = -10.5*1/14 = -0.75 cm

So orientation is pointing down and height of the image is 0.75 cm

So the correct answer is pointing down, height = 0.75 cm

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