The left end of a long glass rod 6.00cm in diameter has a convex

AI Thread Summary
The discussion revolves around determining the image position for a convex hemispherical glass rod with a specific diameter and refractive index. For an object placed infinitely far from the vertex, the expected image position is 13.7 cm, while placing it 12 cm away yields an image position of 8 cm, which was mistakenly identified as the answer for the first part. The participants emphasize the importance of understanding par-axial geometric optics and suggest using ray transfer matrices for complex systems. Clarifications on the expected outcomes for parallel rays and the differences in answers highlight the nuances of optical calculations. Understanding these principles is crucial for accurate image positioning in optical systems.
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The left end of a long glass rod 6.00cm in diameter has a convex hemispherical surface 3.00cm in radius. The refractive index of the glass is 1.60. Distances are measured from the vertex of the hemispherical surface.

A.)Determine the position of the image if an object is placed in air on the axis of the rod at the infinitely far distance to the left of the vertex of the curved end.

I did not know where to start here so I proceeded to B.

B.)Determine the position of the image if an object is placed in air on the axis of the rod at the distance of 12.00cm to the left of the vertex of the curved end.

Here I took the equation (n1/s)+(n2/s')=(n2-n1)/R and solved for s'=n2Rs/((n2-n1)-R) and plugged in 12.00cm in for s and solved to get 8.00cm. I then accidentally hit that answer for A.) and it was right. The correct answer for this part turns out to be 13.7cm, but I do not know why either is that way.

Thanks, Stephen





The Attempt at a Solution

 
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What is it about the answer you don't understand? Can you think of any reason it shouldn't be that way?
BTW: I'd have used the matrix form for this.
 


I can't think of any reason that part B.) shouldn't be 12, but it is 13.7cm, and I have no idea how A.) is 12. I do not know the matrix form
 


What you are doing is called Par-axial geometric optics ... look through that handout for "ray transfer matrix".

At x=0 you have a spherical interface between n=1 (x<0) and n=1.6 (x>0) with R=3.0cm ... we don't care about the situation for |y|>3cm... in fact, we are only using the bit of the system close to the axis (par-axial).

For passage through a spherical surface, the transfer matrix is:
M=\left (<br /> \begin{array}{cc}<br /> 1 &amp; 0\\<br /> \frac{n_1-n_2}{R} &amp; 1<br /> \end{array}\right )

I'm not sure there is a special advantage to this representation for what you need to get. I think you aught to be able to figure it from a ray diagram. The matrix approach is invaluable for systems that are tricky to think about.


Part A wants to know about the situation for parallel rays... what would you expect to happen for this situation?

(aside: I take it you have been given answers of 12cm and 13.7cm respectively?)
 
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