Optics question using ray matrices

[nateastle]

Homework Statement

The problem is: The left end of a long glass rod 10.0 cm in diameter, with an index of refraction 1.5, is ground and polished to a convex hemispherical surface with a radius of 5.0cm. An object in the form of an arrow 2.00mm tall, at right angles to the axis of the rod, is located on the axis 25.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface.

From this I know that R= 5cm, do = 25cm, ho=.2cm, n1 = 1, n2 = 1.5

Homework Equations

n1/n2
-(nlarge - nsmall)/n2R

The Attempt at a Solution

(and what an attempt it is.)
The teacher wants us to use ray matrices to figure out the answer. I figure there are 3 matrices that will need to be formed.

Here is what I have:
$$m1 = \left(\begin{array}{cc}<br /> 1 & 25 \\<br /> 0 & 1\\<br /> \end{array}<br /> \right)$$

$$m2 = \left(\begin{array}{cc}<br /> 1 & 0 \\<br /> -0.067 & .67\\<br /> \end{array}<br /> \right)$$

$$m3 = \left(\begin{array}{cc}<br /> 1 & di \\<br /> 0 & 1\\<br /> \end{array}<br /> \right)$$

When I multiply the 3 together I get my final result as:

$$mfinal = \left(\begin{array}{cc}<br /> 2.675 & 2.675di + 16.75 \\<br /> -.067 & -.067di +.67\\<br /> \end{array}<br /> \right)$$

For some reason I am not getting the right answers in my final matrix to lign up with the image matrix. Any help would be appreciated.

 

[OlderDan]

What order did you use for the matrices when you multiplied them?

 

[nateastle]

I multiplied [m1] [m2] I then took the result and multiplied it by [m3].

 

[OlderDan]

I multiplied [m1] [m2] I then took the result and multiplied it by [m3].

Try [m3][m2][m1] and think about why you might want it that way

 

[nateastle]

I did it that way and that is how I got mFinal. I don't know if my math is not right or if something else is not right.

 

[OlderDan]

I did it that way and that is how I got mFinal. I don't know if my math is not right or if something else is not right.

I think you need to check it. Do you know that [m2][m1] is not the same as [m1][m2]?? I see no way you can have di in the bottom row of the final matrix. I also see no way to get any negatives.

 

[nateastle]

I switched the result of [m2][m1] with where [m3] was suppose to be. Thanks for you help, so here is what I have for my mFinal:
$$mfinal = \left(\begin{array}{cc}<br /> 2.675 + di & 16.75 -.67di\\ <br /> -.067 & .67\\ <br /> \end{array} \right)$$

the negetive I have from my m2 matrix, I posted it wrong orignaly.

 

[OlderDan]

I switched the result of [m2][m1] with where [m3] was suppose to be. Thanks for you help, so here is what I have for my mFinal:
$$mfinal = \left(\begin{array}{cc}<br /> 2.675 + di & 16.75 -.67di\\ <br /> -.067 & .67\\ <br /> \end{array} \right)$$

the negetive I have from my m2 matrix, I posted it wrong orignaly.

The only way I can see you getting a 16.75 anywhere is from multiplying [m1][m2] in that order instead of [m2][m1]. Then [m3][m1][m2] gives your new bottom row but not your top row.

You still need to compute [m3][m2][m1]. You can do it as

{[m3][m2]}[m1] or [m3]{[m2][m1]}

Matrix multiplication is associative, but it is not commutative.

I think you are correct about that sign in [m2]. I assume you changed that back in your original post.