Light Refraction: Finding an Exact Value for a 2nd Image

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Homework Help Overview

The discussion revolves around the topic of light refraction, specifically focusing on finding the exact position of a second image formed by a combination of refracted and reflected light in a system involving a bubble and a concave mirror. Participants are exploring the implications of varying refractive indices and the geometry of the setup.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to determine the position of the first and second images based on given parameters such as refractive indices and distances. Some are questioning how to accurately apply formulas related to image formation and refraction, while others are exploring the implications of the optical path length for reflected light.

Discussion Status

The discussion is ongoing, with various interpretations and calculations being proposed. Some participants have offered specific formulas and methods for finding image positions, while others express uncertainty and seek further clarification from peers.

Contextual Notes

There is mention of the problem being from a junior-level class, indicating a certain complexity. Participants are also navigating the challenge of combining concepts from both refraction and reflection in their analysis.

Shackleford
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One image should be formed by rays that go directly through the bubble and another image is formed by reflected light that hits the bubble.

n1 = 1
n2 = 1.5
R = -7.5 CM
so = 5 cm

If n1 = n2, then si would equal 15 cm as measured from the second vertex on the concave surface. However, they're not equal and the light ray is refracted at the plane. Since n1 is less than n2, the light ray will be refracted away from the perpendicular line. Thus, si' is less than 15 cm.

How do I find an exact value for this second image? I assume the first image is also along the optical axis, but where?

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png
 
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Not sure, but since everything is along the axis, I would think one image appears 5cm*1.5 = 7.5cm away. The other is relected light from the curved end of the lens so that ray has to trave an additional optical path of 2.5*2*1.5 = 7.5cm so that image would appear 7.5cm + 7.5 = 15 cm away, behind the front image.

Youd better hope others will comment on this! :-)
 
rude man said:
Not sure, but since everything is along the axis, I would think one image appears 5cm*1.5 = 7.5cm away. The other is relected light from the curved end of the lens so that ray has to trave an additional optical path of 2.5*2*1.5 = 7.5cm so that image would appear 7.5cm + 7.5 = 15 cm away, behind the front image.

Youd better hope others will comment on this! :-)

How did you formulate your equations?
 
This problem is also from my junior-level class, NOT INTRODUCTORY.
 
For first image, use the formula
refractive index = real depth/apparent depth.
In the problem real depth is 5 cm from the plane surface.Find the apparent depth.
For the second problem, find the position of the image formed by the concave mirror by using the formula
1/u + 1/v = 2/R where u is the object distance, which is 2.5 cm from the pole of the mirror and v is the image distance. Then find the distance of this virtual image from the plane surface. To get the final image use the formula given is the first problem.
 
AD = 5 cm/1.5 = 3.33 cm as measured from the plane interface in the glass hemisphere.

I calculated si = -7.5 cm = v.

This virtual image is 7.5 cm to the right of concave mirror and 15 cm to the right of the plane interface.

Using the original formula, the AD = 10 cm.
 
Last edited:
Your answer is correct.
 

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