Problem 105 - Physics for Scientists and Engineers with Modern Physics

Question

Sam purchases \(+3.50-\mathrm{D}\) eyeglasses which correct his faulty vision to put his near point at 25 \(\mathrm{cm}\) . (Assume he wears the lenses 2.0 \(\mathrm{cm}\) from his eyes.) (a) Calculate the focal length of Sam's glasses. (b) Calculate Sam's near point without glasses. (c) Pam, who has normal eyes with near point at \(25 \mathrm{cm},\) puts on Sam's glasses. Calculate Pam's near point with Sam's glasses on.

Step-by-Step Solution

Answer

(a) The focal length of the glasses is 28.57 cm. (b) Sam's near point without glasses is -117.65 cm. (c) Pam's near point with Sam's glasses is -199 cm.

Step 1: Understand the Formula for Focal Length

The power of the lens is given in diopters, which represents the inverse of the focal length in meters. For eyeglasses with a power of \( P = +3.50 \) diopters, we can calculate the focal length \( f \) using the formula \( f = \frac{1}{P} \). Where \( P \) is the lens power in diopters and \( f \) is the focal length in meters.

Step 2: Calculate the Focal Length

Using the formula from Step 1, substitute \( P = +3.50 \) diopters to find:\[f = \frac{1}{3.50} \approx 0.2857 \text{ meters} = 28.57 \text{ cm}\]

Step 3: Setup the Near Point Equation for Sam

The lens formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o \) is the object distance (Sam's near point without glasses), and \( d_i \) is the image distance (25 cm - 2 cm sight distance). For Sam's glasses, since the image forms at the near point of 25 cm with glasses on, effective distance \( d_i = 23 \) cm.

Step 4: Calculate Sam's Near Point without Glasses

Substitute \( f = 28.57 \) cm and \( d_i = 23 \) cm into the lens formula and solve for \( d_o \). Rearrange the formula to: \[\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i} = \frac{1}{28.57} - \frac{1}{23}\] Calculate: \[\frac{1}{d_o} = 0.0350 - 0.0435 = -0.0085 \, \text{cm}^{-1}\] Thus, \( d_o = \frac{1}{-0.0085} \approx -117.65 \) cm (implying beyond comfortable vision range).

Step 5: Setup the Near Point Equation for Pam

Using the same lens formula, calculate the image distance \( d_i \) for Pam when she wears Sam's glasses. For normal sight, Pam's near point \( d_o \) is 25 cm without glasses. Use lens formula with \( f = 28.57 \) cm: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] Substitute \( d_o = 25 \) cm: \[\frac{1}{28.57} = \frac{1}{25} + \frac{1}{d_i}\]

Step 6: Calculate Pam's Near Point with Sam's Glasses

Rearranging and solving for \( d_i \), Pam's effective distance: \[\frac{1}{d_i} = \frac{1}{28.57} - \frac{1}{25}\] Calculate: \[\frac{1}{d_i} = 0.03498 - 0.04 = -0.00502 \, \text{cm}^{-1}\] Thus, \( d_i = \frac{1}{-0.00502} = -199 \) cm, which indicates that she cannot focus within her typical comfortable range, so the image seems afar.

Key Concepts

Focal Length Calculation

In the world of optics, the focal length is crucial in understanding how lenses affect vision. It's essentially the distance from the center of a lens to its focal point, where light rays converge. Calculating the focal length of eyeglasses is straightforward when the lens's power is known, given in diopters.

• The power of a lens ( P ) is the inverse of its focal length (f).
• Diopters measure lens power and directly tell us the reciprocal focal length in meters.

For Sam's eyeglasses with a power of \(P = +3.50\) diopters, we calculate the focal length using: \(f = \frac{1}{P}\).
Substituting the values, we find the focal length:\[f = \frac{1}{3.50} \approx 0.2857 \text{ meters} = 28.57 \text{ cm}\]This calculation shows that Sam's lenses bring light to a focus 28.57 cm behind the lens.

Lens Formula

The lens formula is an essential tool to determine distances related to lenses for corrected vision. This formula considers the relationship between object distance (\(d_o\)), image distance (\(d_i\)), and focal length (\(f\)):\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]This equation is pivotal in figuring out how Sam's vision is corrected, as well as how it affects Pam when she wears Sam's glasses.
To find Sam's near point without glasses (the point at which he naturally focuses), we rearrange the lens formula:

• Knowing \(f = 28.57\) cm and \(d_i = 23\) cm (effective image distance with glasses), solve for \(d_o\).

This results in:\[\frac{1}{d_o} = \frac{1}{28.57} - \frac{1}{23} = -0.0085 \, \text{cm}^{-1}\]Sam’s natural focus point is thus beyond the comfortable range at:\[d_o \approx -117.65 \, \text{cm}\]Similarly, to compute how Pam's vision changes with Sam's glasses, the process is repeated by inserting Pam's natural near point into the formula, leading us to understand the faulty vision caused by the wrong prescription.

Near Point

The near point is the closest distance at which the eye can focus on an object. For someone with normal vision, this is typically at 25 cm. However, Sam's prescription allows this effect to be adjusted.

• A near point further away implies difficulty focusing on nearby objects, often corrected with lenses.
• A near point too close or negative (calculated as negative distance like Sam's without glasses) indicates objects need to be pulled in extremely close or cannot be comfortably focused.

In Sam's case, the glasses correct his near point to 25 cm effectively. However, Pam, with normal vision, experiences a shift in her near point due to wearing Sam's glasses, resulting in:\[d_i = \frac{1}{-0.00502} = -199 \, \text{cm}\]This situates her near point way beyond her natural comfortable viewing range. Thus, Sam's glasses disrupt the focal point for individuals with normal vision, making it even more essential to have personalized prescription glasses for each individual's needs.