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

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Homework Help Overview

The problem involves a long glass rod with a convex hemispherical surface, focusing on the image formation when an object is placed at varying distances from the vertex of the curved end. The subject area pertains to optics, specifically the behavior of light as it passes through different media and the application of geometric optics principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the lens maker's equation to determine image positions for two different object distances. Some participants question the validity of the results obtained for each part and express confusion regarding the discrepancies in expected outcomes.

Discussion Status

Participants are actively discussing the application of geometric optics principles, with some suggesting the use of matrix methods for analysis. There is an ongoing exploration of the reasons behind the differing image positions and the implications of the par-axial approximation.

Contextual Notes

There is mention of confusion regarding the correct answers for the image positions, with specific values of 12cm and 13.7cm being referenced. The discussion also highlights the importance of understanding the assumptions made in the problem setup, particularly concerning the behavior of light at the spherical interface.

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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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