A lens located in the y-z plane at x = 0 forms an image of an arrow at x = x 2 =

Question

A lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = 106 cm. The tip of the object arrow is located at (x,y) = (x1, y1) = (-54.6 cm, 5.08 cm). The index of refraction of the lens is n = 1.41.

1)What is flens, the focal length of the lens? If the lens is converging flens is positive. It the lens is diverging, flens is negative.

2)What is y2, the y co-ordinate of the image of the tip of the arrow?

3)The lens is a plano-convex lens. What is Rlens, the radius of curvature of the convex side of the lens?

4)The object arrow is now moved to x = x1,new = -16.2 cm. What is x2,new, the new x co-ordinate of the image of the arrow?

Explanation / Answer

Part 1)

Apply 1/f = 1/p + 1/q

1/f = 1/54.6 + 1/106

f = 36.0 cm

Part 2)

y/y' = -q/p

y/5.08 = -(106)/54.6

y = -9.86

Part 3)

Apply 1/f = (n-1)(1/R)

1/36 = (1.41 - 1)(1/R)

R = 14.76 cm

Part 4)

1/36 = 1/16.2 + 1/q

q = -29.5 cm

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