Home > 0877 > GR0877 Problem 15

GR0877 Problem 15

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (E) Recall the lens-maker’s formula: \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right). Here R_1 is the radius of curvature of the lens’ surface closest to the light source and R_2 is the radius of curvature of the lens’ surface farthest from the light source. Sign convention: the radius of curvature is positive if the center of curvature lies to the right from the lens and is negative otherwise. Let the source of light be situated to the left from each of the lenses. Then for (A) one has R_1 = -R, R_2 = R, and f_A \sim -R/2. For (B): R_1 = \infty, R2 = R, f_B \sim -R. (C): R_1 = \infty, R2 = -R, f_C \sim R. For (D) and (E): R_1 = R, R_2 = -R, f \sim R/2. But R_E <R_D, therefore f_E < f_D. Alternatively, lenses (A) and (B) are obviously diverging and from (C), (D), (E) the shortest positive focal length has the one that has the smallest radius of curvature, because it bends a set of parallel rays more than the others, hence, the focal point of this lens is the closest to the lens.

Found a typo? Comment!

Jump to the problem

1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100

Want to get a full PDF with solutions? Read THIS.

The comments appending particular solutions are due to the respective users. Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 2008 by ETS.

Advertisements
Categories: 0877 Tags: , , , ,
  1. Sam
    18.09.2015 at 18:25

    So we have:
    C) F ~ R_C
    E) F ~ R_E/2

    But based on the diagram, since R_C < R_E, this seems ambiguous as to which is smaller. Am I missing something?

  2. Maria
    15.10.2016 at 16:13

    a short focal length does not imply a small radius of curvature

    • 17.10.2016 at 20:46

      It does. One way to see it is to look at the lens-maker’s formula above. Another way would be to take the simplest case — plano-convex lens — and consider an incident ray parallel to its optical axis. Look where the outgoing ray intersects the optical axis (that’s the focal point) if you change the radius of curvature.

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: