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 = 102.1 cm. The tip of the object arrow is located at (x,y) = (x1, y1) = (-52.6 cm, 5.18 cm). The index of refraction of the lens is n = 1.44.

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-coordinate 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 = -16.7 cm. What is x2,new, the new x-coordinate of the image of the arrow?

Explanation / Answer

We know that

1/f = 1/v - 1/u

f = uv / u-v = 52.6 * 102.1/ 154.7 = 34.7cm

The lens is converging

The other cordinate will be

52.6/102.1 = 5.18/h

h = 10.02 cm

Using the lens makers formula

R = f (n - 1)

R = 34.7 (1.44 -1 ) = 15.268 cm

THe new image will be

v = fu /f +u = 34.7 * - 16.7/18 = - 32.2 cm

THe image point will be

32.2/16.7 = 5.18/h

h = 2.65 cm

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