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 = 54.2 cm. The tip of the object arrow is located at (x,y) = (x1, y1) = (-33.2 cm, 6.15 cm). The index of refraction of the lens is n = 1.35.

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.cm

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2)

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

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3)

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

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4)

The object arrow is now moved to x = x1,new = -9.7 cm. What is x2,new, the new x co-ordinate of the image of the arrow?

Explanation / Answer

1) 1/x1 + 1/x2 = 1/F

1/33.2 + 1/54.2 = 1/F

F = +20.58 cm

2) y2/y1 = -x2/x1

y2 = (6.15*54.2)/33.2 = -10.44 cm

3) F = (n-1)(1/R)

R = (n-1)/F = (1.35-1)/20.58 = 0.017 cm

4) 1/9.7 + 1/x2new = 1/20.54

x2new = -18.34 cm

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