Self-Learning Physics Made Simple - Understanding the Eye and Its Imperfections

Physics can be more exciting and accessible than you might think! In this article, we'll delve into a crucial part of the human body: the eye. Instead of just reading dry theory, we'll learn how the eye works through solving specific problems and conducting simple experiments. Often, teachers' efforts to explain concepts or require students to read information can make students passive and unmotivated to engage. This approach will help you gain a deeper understanding of the eye's structure, related optical phenomena, and common vision defects. Let's embark on this scientific journey and make learning physics more engaging and straightforward than ever before!

This article is inspired by the problems presented in the book "Fascinating Problems for Young Physicists" by Nenad Vukmirović and Vladimir Veljić. The book offers a collection of challenging and educational physics problems that provide valuable insights into various physics concepts. By exploring these problems, students can deepen their understanding of physics and develop essential problem-solving skills. You can find more information and access the book here: Fascinating Problems for Young Physicists.

You will gain a thorough understanding of the Eye and significantly enhance your problem-detection and problem-solving skills without having to do any extensive research. Just join me in solving the following simple problem, and you will find that learning Physics becomes more interesting and easier than ever. Let's get started!

Problem on the Structure and Function of the Eye

A schematic view of the structure of the human eye is presented in Figure 1. Light rays that refract at the cornea and eye lens end up at the retina, which produces nerve impulses sent to the brain down the optic nerve. In a simplified model of an eye, the cornea and eye lens can be replaced with one converging lens (called simply the lens in the remainder of the text) while the retina can be modeled as a disk of radius $R = 1.00\ \text{cm}$, the axis of which coincides with the optical axis of the lens, as shown in Figure 2.

Figure 1: Scheme of the structure of the human eye: (1) cornea, (2) eye lens, (3) retina, (4) optic nerve, (5) ciliary muscles, (6) suspensory ligament

The distance between the retina and the lens is $d = 2.40\ \text{cm}$. A human can adjust the focal length of the lens and therefore has the capability of clearly seeing objects at different distances. This process is called eye accommodation and is enabled by ciliary muscles connected to the eye lens by a suspensory ligament. These muscles act to tighten or relax the ligaments and therefore thin down or thicken the lens. Consequently the focal length of the lens changes.

Figure 2: A simplified model of the human eye

1. A human has regular eyesight if images of all objects from a distance larger than $d_0 = 25.0\ \text{cm}$ can be formed at the retina. What is the range of the lens’ focal lengths for a human with regular eyesight?

2. The maximal focal length fmax of the lens for a nearsighted man is smaller than the upper limit of the range determined in part (1). This man uses glasses with a diopter value of $D_1 = −1.00\ \text{m}^{−1}$ to clearly see very distant objects. Determine fmax and find the maximal distance of an object that this man can clearly see without using the glasses. For simplicity neglect the distance between the glasses and the lenses.

3. The minimal focal length fmin of the lens for a farsighted woman is larger than the lower limit of the range determined in part (1). This woman needs glasses with a diopter value of $D_2 = 2.00\ \text{m}^{−1}$ to clearly see objects at a distance of $d_0 = 25.0\ \text{cm}$. Determine fmin and find the minimal distance of an object that this woman can clearly see without using the glasses.

4. A person is nearsighted (farsighted) as well when the distance between the retina and the lens is larger (smaller) than the regular distance of $d = 2.40\ \text{cm}$. Calculate the diopter value of the glasses that should be used by a man with a distance between the retina and the lens of $d_1 = 2.50\ \text{cm}$ $\left(d_2 = 2.30\ \text{cm}\right)$.


Figure 3: Eyes see objects when light from the object reaches the eye

5. A man with regular eyesight whose height is $h = 2.00\ \text{m}$ is observing a tree of height $H = 2h$ (Figure 3). His view is directed toward the middle of the tree. What is the minimal distance between the man and the tree that allows him to see the whole tree?

   Two types of light receptors are placed at the retina – rods (about $N_1 = 10^8$ of them) and cones (about $N_2 = 6·10^6$ of them). Rods enable night vision, while cones are used for vision during the day. Assume that a person can distinguish two distant objects during the day (night) if their images are at different cones (rods). Assume also that the cones (rods) are evenly distributed on the retina surface and that their positions form a square lattice.

6. Two point objects are at a mutual distance of $a = 1.00\ \text{mm}$. The direction that connects them is perpendicular to the optical axis of the lens (Figure 4). What is the maximal distance from which a woman can distinguish between these two objects during the day?


Figure 4: The eye can only distinguish two separate points when the images of these two points are located on two different sensor cells on the retina

7. At what maximal distance can a woman read the license plates of a car during the day? Assume that the license plates can be read if a woman can distinguish between the point objects at a mutual distance of $a = 1.00\ \text{cm}$.

8. At what maximal distance can a woman distinguish between the two eyes of a cat during the night? The eyes of a cat are at a mutual distance of $a = 2.00\ \text{cm}$.

Problem Solution

1. Discover the range of focal length variations of the eye lens

To see an object at a distance $p$ from the eye, a human needs to accommodate the focal length of the lens so that the image of the object is formed at the retina (which is at a distance $l = d$ from the lens). For an object at a distance $p_1 = d_0$ the focal length is given by lens equation $$\frac{1}{f_1}=\frac{1}{p_1}+\frac{1}{l}$$ For an object at a distance $p_2 → ∞$ we have $$\frac{1}{f_2}=\frac{1}{p_2}+\frac{1}{l}$$ From previous equations we obtain $f_1 = 2.19\ \text{cm}$ and $f_2 = 2.40\ \text{cm}$. Consequently the lens focal length of a human with regular eyesight takes a range from $f_1 = 2.19\ \text{cm}$ to $f_2 = 2.40\ \text{cm}$.
With basic geometrical optics knowledge (everyone has learned the thin lens formula since middle school), along with some information from the problem, we have learned a very interesting fact about the eye: The eye is like a special converging lens, its focal length can be changed from $f_1 = 2.19\ \text{cm}$ to $f_2 = 2.40\ \text{cm}$.

2. Calculate the maximum focal length of the eye lens for a nearsighted person and determine the farthest distance they can see clearly

The lens focal length and the distance of the object that the man clearly sees are related by $$\frac{1}{f}=\frac{1}{p}+\frac{1}{l}$$ Consequently, without the use of glasses, this man cannot clearly see objects at a distance larger than $p_\text{max}$, where $$\frac{1}{f_\text{max}}=\frac{1}{p_\text{max}}+\frac{1}{l}\tag{1}$$ The focal length of the system lenses-glasses fns satisfies the relation $$\frac{1}{f_\text{ns}}=\frac{1}{f}+D_1$$ When this man clearly sees very distant objects with the use of glasses, the lens equation reads $$\frac{1}{f_\text{max}}+D_1=\frac{1}{p_2}+\frac{1}{d}\tag{2}$$ where $p_2 → ∞$. From equation (2) we obtain $f_\text{max} = \frac{d}{1−dD_1}= 2.34\ \text{cm}$. By subtracting equations (1) and (2) we find $p_\text{max} = -\frac{1}{D_1} = 1.00\ \text{m}$.
If we know the prescription strength of a nearsighted person's glasses, we can determine the maximum focal length of their eye lens. Additionally, we can find out the interesting fact about the farthest distance at which they can see clearly.

3. Determine the minimum focal length of the eye lens for a farsighted person and the nearest distance they can see clearly without glasses

Without the use of glasses, this woman cannot clearly see objects at a distance smaller than $p_\text{min}$, where $$\frac{1}{f_\text{min}}=\frac{1}{p_\text{min}}+\frac{1}{l}\tag{3}$$ The lens equation for a woman with glasses looking at an object at a distance $d_0$ reads $$\frac{1}{f_\text{min}}+D_2 =\frac{1}{d_0}+\frac{1}{d}\tag{4}$$ From equation (4) it follows that $$f_\text{min}=\frac{1}{\frac{1}{d_0}+\frac{1}{d}-D_2}=2.29\ \text{cm}\tag{5}$$ By subtracting equations (3) and (4) we obtain $$p_\text{min}=\frac{d_0}{1-D_2d_0}=50.0\ \text{cm}\tag{6}$$ By knowing that a farsighted person uses glasses with a diopter strength of $D_2 = 2.00\ \text{dp}$ to see objects clearly at a distance of $d_0 = 25.0\ \text{cm}$, we can determine the minimum focal length of their eye lens. Additionally, we can find out the nearest distance at which they can see clearly without glasses.

4. Calculate the lens power needed for nearsighted and farsighted individuals to see distant objects clearly

The lens equation for a man with regular distance between the lens and the retina when he clearly sees an object at a distance $p$ is $$\frac{1}{f}=\frac{1}{p}+\frac{1}{d}$$ For a man with distance $d_i$ between the retina and the lens who uses glasses with diopter value Di and clearly sees the same object when the lens focal length is the same, we obtain $$\frac{1}{f}+D=\frac{1}{p}+\frac{1}{d_i}$$ Subtracting the previous two equations, we find $$D_i=\frac{1}{d_i}-\frac{1}{d}$$ Consequently, we find in the first case $D_1 = −1.67\ \text{dp}$ and in the second case $D_2 = 1.81\ \text{dp}$.
By knowing the positions of the images for both nearsighted and farsighted individuals when observing distant objects, we can calculate the necessary lens power. For a nearsighted person whose image forms $0.1\ \text{cm}$ in front of the retina $\left(d_1' = 2.3\ \text{cm}\right)$ and a farsighted person whose image forms $0.1\ \text{cm}$ behind the retina $\left(d_2' = 2.5\ \text{cm}\right)$, we can determine the diopter strength of the corrective lenses needed for each to see distant objects clearly.

6. Calculate the minimum distance for a person with normal vision to see the entire tree

A man sees the whole tree when the size $L$ of the image of the tree on the retina is smaller than the retina diameter (Figure 5). Using the similarity of the triangles in Figure 5 we obtain $$\frac{L}{H}=\frac{d}{x}$$ where $x$ is the distance between the man and the tree. Consequently the man sees the whole tree when $$L=H\frac{d}{x}\lt 2R$$ leading to $$x\gt\frac{Hd}{2R}=4.80\ \text{m}$$

Figure 5: To see the entire tree trunk, light emitted from every point of that tree must reach the eye's retina

By understanding the relationship between the observer's height and the height of the tree, we can determine the minimum distance required for the person to see the entire tree. Given that the observer has normal vision and their height is $h = 2.00\ \text{m}$ while the tree's height is $H = 2h$, we can calculate the necessary minimum distance to ensure the entire tree is within their field of view.

6. Determine the maximum distance for the eye to distinguish two points 1.00 mm apart during daylight

The number of cones per unit surface is equal to $$N_S=\frac{N_2}{R^2\pi}$$ On the other hand, since we assume that the positions of cones form a square lattice with lattice constant b, we also have $$N_S=\frac{1}{b^2}$$ From the previous two equations it follows that $$b=R\sqrt{\frac{\pi}{N_2}}=7.24\ \text{μm}$$ When the woman is at a maximal distance at which she can still distinguish between the two objects, the images of the objects are formed at two neighboring cones. From the similarity of triangles in Figure 6, we find \begin{align} \frac{a}{x}&=\frac{b}{d}\\ x&=\frac{ad}{b}=3.32\ \text{m} \end{align}

Figure 6: To distinguish two separate points, the images of those two points created by the eye lens must lie on two separate photoreceptors on the retina.

To determine how far apart two points, separated by $1.00\ \text{mm}$, can be from the eye while still being distinguishable during the day, we use the concepts of visual acuity and the resolving power of the eye. By applying these principles, we can calculate the maximum distance from the eye to these points where they remain discernible, ensuring that they do not merge into a single image under daylight conditions.

7. Calculate the maximum distance to read a car license plate during daylight

From the solution of part (6) we have $$x=\frac{ad}{b}$$ where in this case $a = 1.00\ \text{cm}$, leading to $x = 33.2\ \text{m}$.
To determine the maximum distance at which a person can read a car license plate during the day, given that they need to distinguish objects separated by $1.00\ \text{cm}$, we apply similar principles used in the previous example. By calculating the resolving power of the eye and the visual acuity required to distinguish between these points, we can establish the maximum readable distance for the license plate under daylight conditions.

8. Calculate the maximum distance to distinguish between a cat's eyes at night

Since the woman observes the cat during the night, the solution of part (6) is modified only by replacing the number of cones with the number of rods. Consequently, $$x=\frac{ad\sqrt{N_1}}{R\sqrt{\pi}}=271\ \text{m}$$ To determine the maximum distance at which a person can distinguish between a cat's eyes during the night, given that the eyes are $2.00\ \text{cm}$ apart, we use principles similar to those used for visual acuity calculations in previous examples. By applying these principles, we find that the maximum distance at which a person can distinguish between the cat's eyes in low light conditions is approximately $271\ \text{m}$.

Trong loạt bài toán này, chúng ta đã khám phá nhiều khía cạnh của quang học và thị giác, từ việc xác định tiêu cự và độ tụ của thấu kính cho đến việc tính toán khoảng cách tối đa để phân biệt các vật dưới các điều kiện ánh sáng khác nhau. Mỗi bài toán đều làm nổi bật các nguyên lý quan trọng của vật lý, giúp hiểu cách hệ thống thị giác của chúng ta tương tác với thế giới xung quanh.

Đối với những ai muốn tìm hiểu sâu hơn về các chủ đề thú vị này và khám phá nhiều bài toán tương tự, tôi rất khuyến khích bạn xem quyển sách "Fascinating Problems for Young Physicists." Quyển sách này cung cấp một kho tàng các bài toán thách thức và hấp dẫn trong nhiều lĩnh vực vật lý khác nhau, mang đến cái nhìn sâu sắc và cơ hội thực hành cho học sinh và những người đam mê. Bạn có thể tìm thêm thông tin và truy cập quyển sách tại đây: Fascinating Problems for Young Physicists.

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Chuyên mục: Self-study physics,