Problem 19 - College Physics

Question

Crystalline lens of the eye. The crystalline lens of the eye is double convex and has a typical index of refraction of \(1.43 .\) At minimum power, the front surface has a radius of 10.0 \(\mathrm{mm}\) and the back surface has a radius of 6.0 \(\mathrm{mm}\) ; at maximum power, these radii are 6.0 \(\mathrm{mm}\) and 5.5 \(\mathrm{mm}\) , respectively (although the values do vary from person to person). (a) Find the maximum and minimum power (in diopters) of the crystalline lens if it were in air. (b) What is the range of focal lengths the eye can achieve? (c) At minimum power, where does it focus the image of a very distant object? (d) At maximum power, where does it focus the image of an object at the near point of 25 \(\mathrm{cm} ?\)

Step-by-Step Solution

Answer

The powers are 11.467 D and 15.000 D. The focal lengths range from 66.7 mm to 87.3 mm. At minimum power, it focuses at 87.3 mm; at maximum power, it focuses at 90.9 mm.

Step 1: Understanding the Relevant Formula

The power of a lens is given by the Lensmaker's Formula:\[ P = \left(n - 1\right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]where \(P\) is the power in diopters, \(n\) is the index of refraction, \(R_1\) is the radius of curvature of the front surface, and \(R_2\) is the radius of curvature of the back surface (negative if the surface is concave). For a lens in air, which is our case, \(n = 1.43\).

Step 2: Calculating Minimum Power

Using the radii at minimum power: \(R_1 = 10.0\, \text{mm} = 0.010\, \text{m}\) and \(R_2 = -6.0\, \text{mm} = -0.006\, \text{m}\), we substitute into the formula:\[ P_{\text{min}} = (1.43 - 1) \left( \frac{1}{0.010} - \frac{1}{-0.006} \right) \]Calculate to find: \[ P_{\text{min}} = 43 \times (100 + 166.67) = 43 \times 266.67 = 11.467\, \text{diopters} \]

Step 3: Calculating Maximum Power

For maximum power, use \(R_1 = 6.0\, \text{mm} = 0.006\, \text{m}\) and \(R_2 = -5.5\, \text{mm} = -0.0055\, \text{m}\):\[ P_{\text{max}} = (1.43 - 1) \left( \frac{1}{0.006} - \frac{1}{-0.0055} \right) \]Calculate:\[ P_{\text{max}} = 43 \times (166.67 + 181.82) = 43 \times 348.49 = 15.000\, \text{diopters} \]

Step 4: Calculating Focal Length Range

The focal length \(f\) is the reciprocal of the power:\[ f_{\text{min}} = \frac{1}{P_{\text{min}}} = \frac{1}{11.467} \approx 0.0873\, \text{m} = 87.3\, \text{mm} \]\[ f_{\text{max}} = \frac{1}{P_{\text{max}}} = \frac{1}{15.000} = 0.0667\, \text{m} = 66.7\, \text{mm} \]Therefore, the range of focal lengths is from 66.7 mm to 87.3 mm.

Step 5: Minimum Power Focus for Distant Object

For a very distant object, the object's distance can be considered infinite, making the calculation simpler. The lens at minimum power focuses light at its focal point:\[ s' = f_{\text{min}} \approx 87.3\, \text{mm} \]

Step 6: Maximum Power Focus for Near Point Object

Using the lens formula \(\frac{1}{f} = \frac{1}{s} + \frac{1}{s'}\), with \(s = 250\, \text{mm} = 0.25\, \text{m}\) and \(f = 66.7\, \text{mm} = 0.0667\, \text{m}\), solve for \(s'\):\[ \frac{1}{66.7} = \frac{1}{250} + \frac{1}{s'} \]\[ \frac{1}{s'} = \frac{1}{66.7} - \frac{1}{250} \]Calculate to find:\[ s' \approx 0.0909\, \text{m} = 90.9\, \text{mm} \]

Key Concepts

Crystalline Lens

Our eyes contain a crucial component known as the crystalline lens. This lens is a clear, flexible structure that focuses light onto the retina, helping us see clearly. It behaves as a natural focusing device, adjusting its shape to focus on objects at varying distances. This ability is called accommodation. The lens is double-convex, meaning both surfaces curve outward, which allows it to bend light effectively.
The crystalline lens' shape and composition are key to its function. It is made of water and proteins that align neatly, allowing light to pass through. Unlike other lenses, the crystalline lens can change shape, thanks to the ciliary muscles around it.

• These muscles help the lens become thicker for close objects.
• They allow it to become thinner to see objects far away.

Studying the crystalline lens helps us understand vision disorders like cataracts, where lens proteins become cloudy, blocking light and affecting vision.

Index of Refraction

The index of refraction is a critical concept in understanding how lenses, like the crystalline lens, work. It measures how much light bends as it passes through a medium like glass, water, or our crystalline lens. This property is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium.
For the eye's crystalline lens, the index of refraction ( = 1.43") tells us how effectively it can bend light to help us see.

• An index greater than 1 indicates that light slows down more when passing through the medium than through air.
• In the crystalline lens, the light bends significantly, allowing for the focus on the retina.

The value of the index of refraction is essential for calculating the power of lenses, which in turn determines how they focus light.

Diopters

Diopters are the units used to measure the optical power of a lens, such as the crystalline lens of the eye. They indicate the lens' ability to converge or diverge light. A lens with more diopters offers greater strengthening power, allowing it to focus light from farther or nearer objects effectively.
When calculating the power of any lens, we use the formula:
\[ P = rac{1}{f} \]
where \( P \) is the power in diopters and \( f \) is the focal length in meters.

• The crystalline lens has powers ranging from 11.467 diopters at minimum power to 15 diopters at maximum power.
• Higher diopter values relate to a shorter focal length, allowing the lens to focus on closer objects.

This unit is crucial as it helps ophthalmologists prescribe corrective lenses to compensate for vision issues by modifying the focusing power of the eye.

Focal Length

The focal length of a lens is the distance between the lens and its focus point, where light rays converge. It defines how the lens bends incoming parallel light rays. For the crystalline lens, calculating the focal length range helps us understand how it modifies to focus on objects at different distances.
Using the Lensmaker's formula, we find:

• At minimum power, the lens has a focal length of 87.3 mm.
• At maximum power, it shortens to 66.7 mm.

This range signifies the lens' adaptability, allowing it to change focus from distant to near objects. Longer focal lengths mean the lens stretches to focus on far-away objects, while shorter focal lengths indicate a more significant curvature for nearby objects. Understanding this helps us comprehend how the eye naturally adjusts to maintain clear vision.