Figure P23.59 shows a converging lens with radii R1 = 9.00 cm and R2 = -11.0 cm,

Question

Figure P23.59 shows a converging lens with radii R1 = 9.00 cm and R2 = -11.0 cm, in front of a concave spherical mirror of radius R = 8.00 cm. The focal points (F1 and F2) for the thin lens and the center of curvature (C) of the mirror are also shown. uploaded image Figure P23.60

(a) If the focal points F1 and F2 are 5.00 cm from the vertex of the thin lens, determine the index of refraction for the lens. n = (b) If the lens and mirror are 20.0 cm apart, and an object is placed 8.00 cm to the left of the lens, determine the position of the final image and its magnification as seen by the eye in the figure. The final image is located cm , and the overall magnification is M = multiplied by (c) Is the final image inverted or upright?

Explanation / Answer

(a) 1/f = (n-1) (1/R1 - 1/R2)

1/5 = (n-1) (1/9 - 1/-11)

0.2 = (n-1) *0.20202

n = 1.99

(b) first image distance

i = of / (o-f) = 8*5/(8-5) = 13.333 cm

second object distance 20 - 13.333 = 6.6666 cm

second focal length = R/2 = 4.00

second image distance i = of / (o-f) = 6.666*4.00 / (6.666-4.00) =

= 10.0 cm

So the final image is located 10 cm to the left of the mirror

which is 10 cm to the right of the lens

which is halfway between the lens and mirror

and

M = i1 i2 / o1 o2 = 13.333*10.0 / 8 * 6.6666 = 2.5

(c) the overall magnification is positive, so the image is upright

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