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Question 108:

a. Consider the lens in the human eye, with retina as the right reference plane. In an adult, the distance R is about 0.025 meters (about 1 inch). The ciliary muscles allow you to vary the shape of the lens and thus the lens constant k, within a certain range. What value of k enables you to focus parallel incoming rays, as shown in the figure? This vale of k will allow you to see a distant object clearly. (The customary unit measurement for k is 1 diopter = 1/meter.)

Hint: In terms of transformation $\mathbf{\left[}\begin{array}{l}\mathbf{x}\\ \mathbf{m}\end{array}\mathbf{\right]}\mathbf{→}\mathbf{\left[}\begin{array}{l}\mathbf{y}\\ \mathbf{n}\end{array}\mathbf{\right]}$, you want y to independent of x (y must depend on the slope m alone). Explain why $\frac{\mathbf{1}}{\mathbf{k}}$is called the focal length of the lens.

b. What value of k enables you to read this text from a distance of L = 0.3 meters? Consider the following figure (which is not to scale).

c. An astronomical telescope consists of two lenses with the same optical axis.

Find the matrix of the transformation

$\mathbf{\left[}\begin{array}{l}\mathbf{x}\\ \mathbf{m}\end{array}\mathbf{\right]}\mathbf{→}\mathbf{\left[}\begin{array}{l}\mathbf{y}\\ \mathbf{n}\end{array}\mathbf{\right]}$, in terms of ${\mathit{k}}_{\mathbf{1}}\mathbf{,}{\mathit{k}}_{\mathbf{2}}\mathbf{,}$ and D. For given values of k₁and k₂, how do you choose D so that parallel incoming rays are converted into parallel outgoing rays? What is the relationship between D and focal lengths of the two lenses,$\frac{\mathbf{1}}{{\mathbf{k}}_{\mathbf{1}}}$and $\frac{\mathbf{1}}{{\mathbf{k}}_{\mathbf{2}}}$?

Diagrams -

a. -

b -

c -

Step by step solution

01

Given that

Human eye and telescope

02

Find value of k that focus all parallel rays

As per the hint the transformation matrix for $\left[\begin{array}{l}x\\ m\end{array}\right]→\left[\begin{array}{l}y\\ n\end{array}\right]$is

$\left[\begin{array}{cc}1−Rk& L+R−kLR\\ −k& 1−kL\end{array}\right]$

If the rays ate parallel, then y must depend only on ' m '

This implies 1 - Rk = 0

Thus

$\begin{array}{c}Rk=1\\ ⇒k=\frac{1}{R}\end{array}$

Therefore the value of k is is $\frac{1}{R}$

03

Find values of k that enables to read text from a distance L = 0.3 meters

To focus at one fixed length, the transformation matrix does not depend upon the slope of incoming rays.

This implies $L+R−kLR=0$.

So

$\begin{array}{c}k=\frac{L+R}{LR}\\ =\frac{0.3+0.025}{0.3\left(0.025\right)}\\ =43.33\text{″¾}\end{array}$

Thus,$k=43.33\text{″¾}$

04

Find transformation matrix and relationship between D and focal lengths

For left lens, transformation matrix is same as part (a).

So $k_1=\frac{1}{R_1}$

This implies

$\left[\begin{array}{cc}0& \frac{1}{k_1}\\ −k_1& 1\end{array}\right]$

For right lens

$\begin{array}{l}{L}_1=D−\frac{1}{k_1}\\ ⇒R_1=0\end{array}$

So the transformation matrix becomes becomes:

$\left[\begin{array}{cc}1& D−\frac{1}{k_1}\\ −k_2& 1−k_2\left(D−\frac{1}{k_1}\right)\end{array}\right]$

So, the matrix for both the lenses is

$\begin{array}{c}\left[\begin{array}{cc}0& \frac{1}{k_1}\\ −k_1& 1\end{array}\right]\left[\begin{array}{cc}1& D−\frac{1}{k_1}\\ −k_2& 1−k_{2}D+\frac{k_2}{k_1}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{−k_2}{k_1}& \frac{1}{k_1}−\frac{k_2}{k_1}D+k_2\\ −\left(k_1+k_2\right)& −\left(k_1+k_2\right)D+\frac{k_2}{k_1}\end{array}\right]\end{array}$

Now, to convert parallel incoming rays into outgoing parallel rays, D should be equal to sum of focal lengths of two lenses.

Thus $D=\frac{1}{k_1}+\frac{1}{k_2}$

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