hw6_solution

# hw6_solution - Note that q =-2.79 M = h’/h = 2.1 =-q/p...

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Physics 3C Webassign Homework 6 Solutions: f = R/2 = 10.05cm a. (1/q) = (1/f) – (1/p) = (1./10.05cm) – (1/39.3cm) => q = 13.5cm (real) M = -q/p = -13.5/39.3 = -0.344 (inverted) b. same procedure, q=19.4 (real), M = -0.935 (inverted) c. p = f => q = infinity, magnification is also infinite. No image formed.

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a. 1/q = 1/32 – 1/79 => q = 53.8 and M = -q/p = -53.8/79 = -0.681 (diminished, inverted, real) b. 1/q = 1/32 – 1/20 => q = -53.3 and M = -q/p = 53.8/20 = 2.67 (larger, upright, virtual) a. M = h’/h = 4cm/12cm = 0.33 = -q/p => q = -0.33p (negative, must be virtual) But M>0 so the image is upright. Only convex mirrors produce smaller upright vertical images. b. We know that 37 = p + |q| = p + 0.33p = 1.33p => p= 27.8 cm. c. we know that 1/f = 1/p + 1/q = 1/27.8 – 1/(0.33*27.8) => f = -13.9cm

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Unformatted text preview: Note that q = -2.79 M = h’/h = 2.1 = -q/p => p = -q/2.1 = +2.79cm/2.1 = 1.33 cm. 1/f = 1/q + 1/p = -1/2.79 + 1/1.33 f = 2.54 cm. For a distant object, the light comes in parallel to the axis and is focused through the focal point. Therefore, initially, the film must be at the focal point (q1 = 66 mm). Now, the object position is at p = 2.3m = 2300mm: 1/q = 1/f – 1/p = 1/66mm – 1/2300mm = 0.014716 => q = 67.95mm (farther than q1) So the lens must be moved away from the film by a distance q – q1 = 1.95mm a. 1/q = 1/f -1/p = 1/21. – 1/39.2 => q = 45.2cm (real => side of lens opposite object) M = -q/p = -1.15 (inverted). b. 1/q = 1/f – 1/p = 1/21 – 1/21 => q= infinity = M. (no image) c. 1/q = 1/21 – 1/10.3 => q = -20.2 (virtual, same side of lens as object) M = -q/p = 1.96 (upright)....
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hw6_solution - Note that q =-2.79 M = h’/h = 2.1 =-q/p...

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