Please answer Question # 6. The answer to number 6 depends on answer to Question

Question

Please answer Question # 6. The answer to number 6 depends on answer to Question #5· The correct answer to #5 according to grading system is 50.83 cm. Question 5, chap 136, sect 6 part 1 of 2 10 points Answers previously submitted that were wrong: 11.9266, -16.8, 44.2, -22.7,-3.5266 and -8.36 A lens of thickness-8.4 cm has convex radii of curvature of magnitude R1 = 9.8 cm and R2 = 16.8 cm. The index of refrac- tion of the lens is n = 1.56. An object O" is placed 37.8 cm to the left of the lens. Ri Where is , the image due to the first surface, measured from that surface (with positive to the right)? Answer in units of em Question 6, chap 136, sect 6 part 2 of 2 10 points Where is 12, the image due to the second surface, relative to the first surface? Answer in units of m

Explanation / Answer

focal length of a thick lens with radii of curvature R1 and R2 and thickness t is given by
1/f = (n - 1)(1/R1 - 1/R2 + (n-1)t/nR1*R2)
where n is refractive index of the lens and the lens is in air

for the given situation
R1 = 9.8 cm
R2 = -16.8 cm
t = 8.4 cm
n = 1.56
hence
1/f = (1.56 - 1)(1/9.8 + 1/16.8 - (1.56-1)8.4/1.56*9.8*16.8)
f = 6.9808 cm

also, from gaussean refraction formula for refraction at curved surface
for refraction from air to glass
n/v - 1/u = (n - 1)/R1
where v is image distance, u is objkect distance
u = -37.8 cm
hence
1.56/v + 1/37.8 = 0.56/9.8
v = 50.834 cmto the right of left surface

for second surface
u = 50.834-8.4 = 42.434 cm
R2 = -16.8 cm
1.56/v - 1/42.434 = -0.56/16.8
v = -159.7118 cm to the left of the second surface

v' = -151.31181364 cm to the left of the left surface

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