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 = 85 cm. The tip of the object arrow is located at (x,y) = (x1, y1) = (-34.7 cm, 6.08 cm). The index of refraction of the lens is n = 1.33.

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 = -11.1 cm. What is x2,new, the new x co-ordinate of the image of the arrow?

5.Is the new image of the arrow real or virtual? Is it upright or inverted?

Explanation / Answer

x1 = 34.7

x2 = 85

(1/x1) + (1/x2) = (1/f)

(1/34.7)+(1/85) = 1/f

f = 24.64 cm

from lens maker equation 1/f = (n-1)*(1/R)

1/24.64 = (1.33-1)*(1/R)

R = 8.1312 cm <---------asnwer

4)

1/x1new + 1/x2new = 1/f

(1/11.1) + (1/x2new) = 1/24.64

x2 new = -20.199 cm

5) virtual , upright

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