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 = 133.5 cm. The tip of the object arrow is located at (x,y) = (x1, y1) = (-71.9 cm, 6.72 cm). The index of refraction of the lens is n = 1.53.

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 = -29.9 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?

Real and upright

Real and inverted

Virtual and upright

Virtual and inverted

Explanation / Answer

1) from the given data

object distance, u = 71.9 cm

image distance, v = 133.5 cm

let f is the focal length of the lens

Apply, 1/f = 1/u + 1/v

1/f = 1/71.9 + 1/133.5

f = 46.73 cm

It is converging lens.

2) magnification, m = -v/u

= -133.5/71.9

= -1.857

now use, y2 = m*y1

= -1.857*6.72

= -12.48 cm

3) 1/f = (n-1)*(1/infinite + 1/R)

1/f = (n-1)/R

R = (n-1)*f

= (1.53 - 1)*46.73

= 24.77 cm

4) Apply, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/46.73 - 1/29.9

v = -83.02 cm

so, x2 = -83.02 cm

5) virtual and upright

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