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fluidistic
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Homework Statement
A biconconvex (n_l=1.5) lens have radii worth 20 and 10 cm and an axial width of 5 cm. Describe the image of an object whose height is 2.5 cm and situated at 8 cm from the first vertex.
Homework Equations
Transfer matrices.
The Attempt at a Solution
I used the ray transfer matrix method shown in the website http://physics.tamuk.edu/~suson/html/4323/thick.html.
Let [tex]M[/tex] be the ray transfer matrix. Then [tex]M=M_3M_2M_1[/tex].
M_1 being a refraction matrix of the first lens, M_2 the ray transfer matrix of the ray passing through the lens and M_3 the refraction of the ray while leaving the lens through the second radius of the lens.
Mathematically [tex]M_1= \begin{bmatrix}1 & 0 \\ \frac{1}{0.2m} \left ( \frac{1}{1.5}-1 \right ) & \frac{1}{1.5} \end{bmatrix}[/tex].
[tex]M_2=\begin{bmatrix}1 & 0.05 \\ 0 & 1 \end{bmatrix}[/tex].
[tex]M_3= \begin{bmatrix}1 & 0 \\ -\frac{1}{0.1m} \left ( 1.5-1 \right ) & 1.5 \end{bmatrix}[/tex].
And thus M gave me [tex]\begin{bmatrix}1 & \frac{1}{20} \\ 3 & \frac{17}{12} \end{bmatrix}[/tex].
Therefore [tex]\begin{bmatrix} y_1 \\ \theta _1 \end{bmatrix}=\begin{bmatrix}1 & \frac{1}{20} \\ 3 & \frac{17}{12} \end{bmatrix} \begin{bmatrix} y_0 \\ \theta _0 \end{bmatrix} =\begin{bmatrix} \frac{1}{40} \\ -\frac{3}{40} \end{bmatrix}[/tex].
Thus the special ray leaving the maximum height of the object and parallel to the optics axe enter the lens at an height of 1/40 m and leaves the lens at the same height... seems totally counter intuitive to me, therefore wrong.
Anyway it gives me the equation [tex]y(x)=-\frac{3x}{40}+\frac{1}{20}[/tex], solving for y(x)=0, I reach a focal distance of 2/3 m! Way too big in my opinion but I'm not 100% sure. Also the image seems real, which totally disagree with my intuition once again since the object is much closer to the lens than the focal length. In disagreement with http://upload.wikimedia.org/wikipedia/commons/9/97/Lens3b.svg, also I know it's a thin lens.
Any help is greatly appreciated.
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