Problem 35 - University Physics with Modern Physics

Question

The cornea behaves as a thin lens of focal length approximately 1.8 cm, although this varies a bit. The material of which it is made has an index of refraction of 1.38, and its front surface is convex, with a radius of curvature of 5.0 mm. (a) If this focal length is in air, what is the radius of curvature of the back side of the cornea? (b) The closest distance at which a typical person can focus on an object (called the near point) is about 25 cm, although this varies considerably with age. Where would the cornea focus the image of an 8.0-mm-tall object at the near point? (c) What is the height of the image in part (b)? Is this image real or virtual? Is it erect or inverted? (\(Note:\) The results obtained here are not strictly accurate because, on one side, the cornea has a fluid with a refractive index different from that of air.)

Step-by-Step Solution

Answer

(a) R2 ≈ 49.02 mm; (b) Image position: ≈ 1.937 cm; (c) Image height: ≈ -0.62 mm, real, inverted.

Step 1: Understand the Lensmaker's Formula

The lensmaker's formula relates the focal length ( f), refractive index (n), and radii of curvature ( R1 and R2) of a lens: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where R1 is the radius of curvature of the front surface and R2 is the radius of curvature of the back surface. For a thin lens in air, n is 1.38.

Step 2: Apply Lensmaker's Formula for Part (a)

Given f = 1.8 cm (or 18 mm), n = 1.38, and R1 = 5.0 mm, we need to find R2. Rearrange the formula to solve for R2: \[ \frac{1}{R_2} = (n - 1) \frac{1}{R_1} - \frac{1}{f} \] Plug in the values: \[ \frac{1}{R_2} = (1.38 - 1) \frac{1}{5} - \frac{1}{18} \] Solve to find R2.

Step 3: Calculate R2

Calculate the expression: \[ (0.38) \times \frac{1}{5} = 0.076 \] Subtract this from \( \frac{1}{18} \) (which is approximately 0.0556): \[ \frac{1}{R_2} = 0.076 - 0.0556 = 0.0204 \] Therefore, R2 = \( \frac{1}{0.0204} \approx 49.02 \) mm.

Step 4: Find the Image Distance for Part (b)

Use the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) to find d_i (image distance), where \( d_o = 25 \) cm and \( f = 1.8 \) cm.: Rearrange to solve for d_i: \[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \] Plug in the values: \[ \frac{1}{d_i} = \frac{1}{1.8} - \frac{1}{25} \] Calculate \( d_i \).

Step 5: Solve for Image Distance (d_i)

Calculate the expression: \[ \frac{1}{1.8} \approx 0.556 \] \[ \frac{1}{25} \approx 0.04 \] Then, \( \frac{1}{d_i} = 0.556 - 0.04 = 0.516 \) Therefore, \( d_i = \frac{1}{0.516} \approx 1.937 \) cm.

Step 6: Determine Image Height for Part (c)

Use magnification formula \( m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \), where h_o = 8.0 mm and we've calculated d_i = 1.937 cm (convert to mm: 19.37 mm). Rearrange to find h_i: \[ h_i = h_o \times \frac{d_i}{d_o} \] Plug in the values: \[ h_i = 8.0 \times \frac{19.37}{250} \approx -0.62 \] mm.

Step 7: Interpret Image Characteristics

Since h_i is negative, the image is inverted (not erect). A positive d_i greater than the focal length implies a real image, since the object is beyond focal length and di is positive.

Key Concepts

Cornea Refraction

The cornea, part of our eye, plays a crucial role in focusing light. It acts like a lens with a specific focal length and refractive index. The cornea's shape is important because it comprises two curved surfaces, each surface affecting how light is bent or refracted. This bending of light helps us see clearly. The Lensmaker's formula is what we use to understand cornea refraction:\[\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]Here, \(n\) denotes the refractive index of the cornea material, \(R_1\) is the radius of curvature of the front surface, and \(R_2\) is for the back. Knowing these values helps us find out how effectively the cornea can focus light, making it possible to correct vision problems with precision. The refractive index of 1.38 tells us how much the light slows down when it passes through the cornea.
This formula captures how the different curves of the cornea contribute to focusing light onto the retina, just like a camera lens focuses light onto film.

Image Distance Calculation

Determining where an image will form is essential for understanding vision. To find the image distance (\(d_i\)), we use the lens formula:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]This equation comes in handy when we need to figure out image placement by using the known object distance (\(d_o\)), which is the distance from the object to the lens, and the focal length (\(f\)). For instance, if an object is 25 cm away from the cornea, we want to find where its image will appear inside the eye. By rearranging the formula, we can solve for \(d_i\):\[\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}\]Plugging in the values, we can calculate \(d_i\), which informs us whether the image forms on the retina for clear vision or not. This information is critical for diagnosing and correcting vision, as improper image focus can lead to blurry images or the need for glasses.

Magnification Formula

Magnification tells us how large or small an image appears compared to the actual object. In optics, this is quantified using the magnification formula:\[m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\]In this formula, \(h_i\) is the height of the image, while \(h_o\) is the height of the object. The negative sign indicates that the image is inverted relative to the object. Using the formula, we can also determine if the image is real or virtual based on the sign of \(d_i\).
When we calculate \(h_i\), the result shows us how the image size and orientation compare to the actual object. For example, if an image's height is negative, the image is inverted, and if it is positive, it's upright. Such details help optometrists prescribe the right kind of corrective lenses.
Understanding magnification is key in determining whether an image is erect or inverted, helping in both everyday vision correction and advanced optical designs.