Refraction Problem istance Coin Appears to be in Glass Sphere

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Refraction Problem:Distance Coin Appears to be in Glass Sphere

Homework Statement


A solid glass sphere with a radius of 5.00 cm and index of refraction of 1.52 has a small coin embedded 3.00 cm from the front surface of the sphere. For the viewer looking at the coin through the glass, at what distance from the front surface of the glass does the coin's image appear to be located?


Homework Equations


Snell's Law


The Attempt at a Solution


I tried drawing two rays from the center of the coin, one as a diameter of the sphere and another through the top of the sphere, and found that the intersection of the first ray and the refracted second ray was at 1.97 cm from the front of the mirror. But that wasn't the correct answer. It looks like geometric interpretation won't be of use. I also tried Lensmaker's equation with curvature for the second side to be infinity but that also yielded the incorrect answer. How would you approach this?
 

Answers and Replies

  • #2
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I tried it out by another method, and my answer comes out to be 1.973 cm too....Maybe, the given solution is incorrect.
 
  • #3
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May you outline your method?
 
  • #4
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I simply used the equation that gives the refractive index of any substance in terms of apparent depth to actual depth.
 
  • #5
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Hm, 1.97 is not even an answer choice. The answer choices are
a. 2.48 cm
b. 3.20 cm
c. 5.00 cm
d. 6.85 cm

There is a similar problem as follows:

A goldfish is swimming in water (n=1.33) inside a spherical plastic bowl of index of refraction 1.33. If the goldfish is 10 cm from the front wall of the 15-cm radius bowl, where does the goldfish appear to an observer in front of the bowl?

Using your method, the answer comes out as 10/1.33=7.5 which also isn't an answer choice. The answer choices are:
a. 6.0 cm behind the plastic
b. 7.0 cm behind the plastic
c. 8.0 cm behind the plastic
d. 9.0 cm behind the plastic
 
  • #6
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Aha, I just noticed the spherical in the second problem. The formula for it would change accordingly (Today doesn't seem to be a good day for me to solve stuff, it seems :frown:)

Use the equation that relates the object distance, image distance and the radius of the curved surface. If you do not know about it, it can be derived in the same way you derive the equation for plane surface, but with a curved surface of radius R also involved.

PS: I believe the answer to the second one is 9 cm.
 
  • #7
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Oh yes, I was just about to tell you about the other equation ([itex]\frac{n_1}{s}+\frac{n_2}{i}=\frac{n_2-n_1}{R}[/itex]) where I get the correct answer. Thanks! I also get 9.0 cm for the second problem, though the Answer Key says 8.0 cm. I think it's just a typo on their part.
 
  • #8
881
40


Oh yes, I was just about to tell you about the other equation ([itex]\frac{n_1}{s}+\frac{n_2}{i}=\frac{n_2-n_1}{R}[/itex]) where I get the correct answer. Thanks! I also get 9.0 cm for the second problem, though the Answer Key says 8.0 cm. I think it's just a typo on their part.

Yep, thats the one! :smile:
 

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