- #1
frankR
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Here is the problem:
A glass sphere with a diameter of 5cm has a scratch on its surface. When the scratch is viewed through the glass from a position directly opposite, where is the virtual image of the scratch, and its magnification? The glass has an index of refraction n=1.50. Explain the result.
[Answer: s'=-10cm; 3x]
I've constructed a system of matrices that looks like this:
M3*M2*M1
M1 is for the first refraction of the scratch at the sphere's front surface.
M2 is for the translation of the ray through the glass.
M3 is for the second refraction through the second surface of the glass where the viewer is located.
When I multiply the matrices together I get these equations.
yf=-2/3yo + 10/3αo
αf=-4/15yo + 1/3αo
I do not understand how I use the geometry of the ray to locate an image of a scratch. What am I missing in all of this? Did I use the matrix equation properly? What do I do, I'm extremely stumped?
Thanks for any help.
Frank
A glass sphere with a diameter of 5cm has a scratch on its surface. When the scratch is viewed through the glass from a position directly opposite, where is the virtual image of the scratch, and its magnification? The glass has an index of refraction n=1.50. Explain the result.
[Answer: s'=-10cm; 3x]
I've constructed a system of matrices that looks like this:
M3*M2*M1
M1 is for the first refraction of the scratch at the sphere's front surface.
M2 is for the translation of the ray through the glass.
M3 is for the second refraction through the second surface of the glass where the viewer is located.
When I multiply the matrices together I get these equations.
yf=-2/3yo + 10/3αo
αf=-4/15yo + 1/3αo
I do not understand how I use the geometry of the ray to locate an image of a scratch. What am I missing in all of this? Did I use the matrix equation properly? What do I do, I'm extremely stumped?
Thanks for any help.
Frank