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

AI Thread Summary
The discussion focuses on calculating the position of a second image formed by light refraction and reflection in a bubble. The refractive indices and curvature of the bubble are considered, with the first image appearing at 5 cm and the second image calculated to be 15 cm behind the first due to additional optical path length. The participants discuss using formulas related to refractive index and mirror equations to determine image distances. The final calculations confirm that the virtual image is positioned 7.5 cm to the right of the concave mirror and 15 cm from the plane interface. The conversation emphasizes the importance of understanding light behavior in optical systems.
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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