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

[sckeen1988]

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

 

[Simon Bridge]

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.

 

[sckeen1988]

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

 

[Simon Bridge]

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:
[tex]M=\left (
\begin{array}{cc}
1 & 0\\
\frac{n_1-n_2}{R} & 1
\end{array}\right )[/tex]

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?)

 

[Nickie]

Hi Stephen,

It seems like you have made a small mistake in your calculation for part B. The correct answer for the position of the image in this case should be 13.7cm, as you mentioned. This can be found by using the same equation you used for part A, but plugging in the distance of 12.00cm for s instead of the infinitely far distance. The reason for this difference is because as the object gets closer to the hemispherical surface, the image will also move closer to the surface, resulting in a shorter distance between the object and the image.

In terms of understanding the concept behind this calculation, it is helpful to think about how light behaves when passing through different mediums. In this case, the light is traveling from air (with a refractive index of 1.00) to glass (with a refractive index of 1.60). When the light hits the curved surface of the glass, it is bent or refracted. This bending of light is what creates the image of the object. The amount of bending depends on the curvature of the surface and the refractive index of the material. In this case, the refractive index of the glass is higher than that of air, causing the light to bend more, and resulting in a shorter distance between the object and the image.

I hope this helps to clarify your understanding. Keep up the good work in your scientific studies!