Problem 25 - College Physics

Question

A student's far point is at \(17.0 \mathrm{cm},\) and she needs glasses to view her computer screen comfortably at a distance of 45.0 \(\mathrm{cm} .\) What should be the power of the lenses for her glasses?

Step-by-Step Solution

Answer

The lens power should be -0.81 diopters.

Step 1: Understand the Problem

The student's far point is 17 cm, which means she cannot see objects clearly beyond this distance. The goal is to calculate the power of the lenses needed so that she can see clearly at 45 cm, which is the desired distance to the computer screen.

Step 2: Use the Lens Formula

To find the required lens power, we use the formula for lens power: \[ P = \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where \(f\) is the focal length, \(d_o\) is the distance of the object (computer screen), and \(d_i\) is the distance to the image (the student’s far point).

Step 3: Substitute the Values

Assign the values \(d_o = -45.0\, \mathrm{cm}\) (it's taken as negative because the object is virtual with respect to the lens), and \(d_i = -17.0\, \mathrm{cm}\) (negative because the far point is beyond the eye indicating virtual image for myopic correction): \[ \frac{1}{f} = \frac{1}{-45.0} + \frac{1}{-17.0} \]

Step 4: Calculate the Inverse Focal Length

Calculate the result of the fractions: \[ \frac{1}{f} = \frac{-1}{45.0} + \frac{-1}{17.0} = \frac{-1 \times 17 + -1 \times 45}{45 \times 17} = \frac{-17 - 45}{765} = \frac{-62}{765} \]

Step 5: Solve for Lens Power (P)

The power of the lens \(P\) is the inverse of the focal length (in meters): \[ P = \frac{1}{f} = \frac{-62}{765} \, \text{cm}^{-1} = -0.081 \, \text{cm}^{-1} \]Convert to diopters (1 D = 100 cm):\[ P = -0.081 \, \text{cm}^{-1} \times 100 = -0.81 \, \text{D} \]

Step 6: Final Answer

The power needed for the lenses is \(-0.81\) diopters, meaning the lenses should be diverging (negative power) to correct the myopia.

Key Concepts

Lens Formula

The lens formula is a key concept in physics, specifically in optics, which deals with the bending of light as it passes through lenses to form images. The formula is expressed as:\[ P = \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Here's what each variable means:

• \(P\) is the power of the lens, measured in diopters (D). It tells us how effectively a lens can converge (positive power) or diverge (negative power) light.
• \(f\) is the focal length of the lens, the distance from the lens to the point where it focuses light. The focal length is in centimeters or meters.
• \(d_o\) stands for the distance between the object and the lens.
• \(d_i\) represents the distance between the image and the lens.

The power of the lens is a real-world application that helps correct vision problems like myopia.To solve problems using the lens formula, substituting known values and solving for the unknown is key. Negative signs indicate virtual images or diverging lenses, which are crucial concepts to understand when dealing with myopia correction.

Myopia Correction

Myopia, or nearsightedness, is a common vision condition where distant objects appear blurry, while close ones remain clear. This occurs when the eyeball is too long, or the cornea is too curved. Light focuses in front of the retina instead of directly on it. Fortunately, optical lenses can correct this condition effectively. The most common method of correction involves the use of diverging lenses, which have a negative power. These lenses spread out light rays before they reach the eye, moving the focal point back onto the retina, thus allowing clear vision of distant objects. In the exercise, the student must see the computer screen clearly from 45 cm. We calculated that the required lens power is -0.81 D, indicating that diverging lenses are necessary. These lenses will adjust the light path to meet the distant point of vision with clarity. Thus, proper lens calculation is crucial for effective vision correction.

Optical Lenses

Optical lenses are transparent objects, often made of glass or plastic, that refract light. We have two main types of lenses used for different purposes in optics:

• Converging Lenses: Also known as convex lenses, they focus incoming light rays to a point. These are used in the correction of hypermetropia (farsightedness), where distant objects are clear, but close ones are not.
• Diverging Lenses: Also known as concave lenses, they spread out light rays, making them appear to originate from a point behind the lens. These lenses help correct myopia (nearsightedness).

To choose the right lens for vision correction, understanding its properties is essential. Diverging lenses' ability to spread light makes them ideal for correcting myopic vision problems. Calculating the correct power, as we have done, ensures that the light is correctly focused onto the retina, allowing for clear vision at requisite distances. Good comprehension of optical lenses and their applications can lead to significant improvements in problem-solving with respect to vision correction.