- #1

frankR

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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:

M

_{3}*M

_{2}*M

_{1}

M

_{1}is for the first refraction of the scratch at the sphere's front surface.

M

_{2}is for the translation of the ray through the glass.

M

_{3}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.

y

_{f}=-2/3y

_{o}+ 10/3α

_{o}

α

_{f}=-4/15y

_{o}+ 1/3α

_{o}

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