Figure P36.72 shows a thin converging lens for which the radii of curvature are

Question

Figure P36.72 shows a thin converging lens for which the radii of curvature are R1 = 9.20 cm and R2 = -11.0 cm. The lens is in front of a concave spherical mirror with the radius of curvature |R| = 9.10 cm. (a) Assume that its focal points F1 and F2 are 5.00 cm from the center of the lens. Determine its index of refraction. (b) 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 (as measured to the left of the lens) and its magnification as seen by the eye in the figure.

Explanation / Answer

(A) Applying lens maker's formula,

1/f = (n - 1) [ 1/R1 - 1/R2]

1/5 = (n - 1) [ 1/9.20 + 1/11]

n - 1 = 1

n = 2

(B) for lens:

object distance, p1 = 8 cm

f1 = 5 cm

Applying lens equation, 1/f =1/p + 1/q

1/5 =1/8 + 1/q1

q1 = 13.33 cm

now this image will work as object for mirror.

p2 = 20 - 13.33 = 6.67 cm

f = 9.10 / 2 = 4.55 cm

1/4.55 = 1/6.67 + 1/q2

q2 = 14.3 cm

now this image will work as object for lens,

p3 = 20 - 14.3 = 5.68 cm

f = 5 cm

1/5 = 1/5.68 + 1/q3

q3 = 41.5 cm left of lens . ..........Ans

Magnification = (-q1/p1)(-q2/p2) (-q3 /p3)

= - ( 13.33 / 8) ( 14.3 / 6.67) (41.5 / 5.68)

= - 26.1 ..........Ans

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