/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 A student is reading material wr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A student is reading material written on a blackboard. Her contact lenses have a refractive power of 57.50 diopters; the lens-to-retina distance is 1.750 cm. (a) How far (in meters) is the blackboard from her eyes? (b) If the material written on the blackboard is 5.00 cm high, what is the size of the image on her retina?

Short Answer

Expert verified
(a) The blackboard is approximately 2.78 meters away. (b) The image size on the retina is calculated using the magnification ratio.

Step by step solution

01

Understand the Refractive Power

The refractive power (P) of the lens is given as 57.50 diopters. The formula for refractive power is \( P = \frac{1}{f} \), where \( f \) is the focal length in meters. We need to calculate \( f \) using this formula.
02

Calculate the Focal Length

First, convert the given refractive power into focal length. We use the formula: \[ f = \frac{1}{P} = \frac{1}{57.50} = 0.01739 \text{ meters.} \]
03

Lens-to-Retina Distance Conversion

The lens-to-retina distance is given as 1.750 cm. Convert this distance into meters by dividing by 100: \[ 1.750 \text{ cm} = 0.0175 \text{ meters} \]
04

Use Lens Formula to Find Object Distance

We use the lens formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] where \( v \) is the image distance (0.0175 m) and \( u \) is the object distance (distance to the blackboard). Substituting the known values:\[ \frac{1}{0.01739} = \frac{1}{0.0175} + \frac{1}{u} \] Solve for \( u \):Calculate \( u \) using this information.
05

Solve for Blackboard Distance

Rearrange the formula to solve for \( u \):\[ \frac{1}{u} = \frac{1}{0.01739} - \frac{1}{0.0175} \] Calculate \( u \):\[ \frac{1}{u} = 57.50 - 57.14 \approx 0.36 \]\[ u \approx \frac{1}{0.36} = 2.78 \text{ meters} \]
06

Magnifying Formula for Image Size

To find the image size \( h' \), use the magnification formula: \[ M = \frac{h'}{h} = \frac{v}{u} \] where \( h = 5.00 \text{ cm} \). Thus, \[ h' = h \times \frac{v}{u} \]
07

Calculate Image Size on Retina

Substitute the known values: \[ h' = 5.00 \times \frac{0.0175}{2.78} \text{ cm} \] Calculate \( h' \) to find the image size on the retina.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refractive Power
Refractive power is an essential concept in optics, and it measures how effectively a lens can bend light. It is expressed in units called diopters. The formula for calculating refractive power (\( P \)) is \( P = \frac{1}{f} \), where \( f \) is the focal length of the lens in meters.
If the refractive power is high, the lens is stronger and can focus light quickly over a shorter distance. Conversely, a lower refractive power indicates a weaker lens. Understanding refractive power can assist in analyzing how corrective lenses, like those used in eyeglasses and contact lenses, help improve vision. In the exercise, the student's lenses have a refractive power of 57.50 diopters, indicating highly focusing lenses.
Lens Formula
The lens formula is a fundamental equation in optics, which relates the object distance (\( u \)), the image distance (\( v \)), and the focal length (\( f \)) of a lens. This formula is expressed as:\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]
This equation helps to find unknown distances, either how far an object is from the lens or where the image will be formed. For instance, the exercise uses the lens formula to locate how far the blackboard is from the student's eyes (\( u \)).
By substituting the known values—focal length and lens-to-retina distance—into the formula, you can solve for \( u \), the object distance. This calculation is pivotal in applications like eyeglasses and camera lenses where precision is critical.
Focal Length
Focal length is the distance between the lens and the focus point where parallel rays of light converge or diverge. It is a critical parameter that affects the performance of lenses in imaging systems. The formula \( f = \frac{1}{P} \) calculates the focal length from the refractive power.
  • A short focal length indicates a strong lens, capable of bending light significantly.
  • Conversely, a long focal length corresponds to a less powerful lens.
For the student's lenses with a refractive power of 57.50 diopters, the calculated focal length is 0.01739 meters. This short focal length indicates that the lenses are quite powerful, which is necessary for correcting certain vision impairments.
Lens-to-Retina Distance
The lens-to-retina distance is an anatomical measure reflecting the distance light travels inside the eye after passing through the lens before it reaches the retina. In general, this distance is very short, because the retina is located at the back of the eye.
In this exercise, the lens-to-retina distance is provided as 1.750 cm, which is equivalent to 0.0175 meters. This conversion is essential for using consistent units in optical calculations like those performed using the lens formula.
This distance helps in determining how images are projected onto the retina, influencing factors such as magnification and image size. These calculations are utilized in predicting how a person's vision will interpret objects at various distances, like reading a blackboard from a classroom seat.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A forensic pathologist is viewing heart muscle cells with a microscope that has two selectable objectives with refracting powers of 100 and 300 diopters. When he uses the 100-diopter objective, the image of a cell subtends an angle of \(3 \times 10^{-3}\) rad with the eye. What angle is subtended when he uses the 300 -diopter objective?

The refractive indices of materials \(A\) and \(B\) have a ratio of \(n_{A} / n_{B}=1.33\) . The speed of light in material \(A\) is \(1.25 \times 10^{8} \mathrm{m} / \mathrm{s}\) . What is the speed of light in material \(B ?\)

Two converging lenses \(\left(f_{1}=9.00 \mathrm{cm} \text { and } f_{2}=6.00 \mathrm{cm}\right)\) are separated by 18.0 \(\mathrm{cm}\) . The lens on the left has the longer focal length. An object stands 12.0 \(\mathrm{cm}\) to the left of the left-hand lens in the combination. (a) Locate the final image relative to the lens on the right. (b) Obtain the overall magnification. (c) Is the final image real or virtual? With respect to the original object, (d) is the final image upright or inverted and (e) is it larger or smaller?

A converging lens has a focal length of 88.00 cm. An object 13.0 cm tall is located 155.0 cm in front of this lens. (a) What is the image distance? (b) Is the image real or virtual? (c) What is the image height? Be sure to include the proper algebraic sign.

A farsighted person can read printing as close as 25.0 cm when she wears contacts that have a focal length of 45.4 cm. One day, she forgets her contacts and uses a magnifying glass, as in Figure 26.39b. Its maximum angular magnification is 7.50 for a young person with a normal near point of 25.0 cm. What is the maximum angular magnification that the magnifying glass can provide for her?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.