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

[Shackleford]

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
s_o = 5 cm

If n₁ = n₂, then s_i 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, s_i^' 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

 

[rude man]

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! :-)

 

[Shackleford]

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?

 

[Shackleford]

This problem is also from my junior-level class, NOT INTRODUCTORY.

 

[rl.bhat]

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.

 

[Shackleford]

AD = 5 cm/1.5 = 3.33 cm as measured from the plane interface in the glass hemisphere.

I calculated s_i = -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.

 

[rl.bhat]

Your answer is correct.