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Chapter 24: Problem 34

A person can see clearly only upto a distance of 30 \(\mathrm{cm}\). He wants to read a book placed at a distance of 50 \(\mathrm{cm}\) from his eyes. What is the power of the lens of his spectacles? (a) \(-1.0 \mathrm{D}\) (b) \(-1.33 \mathrm{D}\) (c) - 1.67 D (d) \(-2.0 \mathrm{D}\)

Short Answer

Expert verified
The power of the lens is -1.33 D, option (b).

Step by step solution

01

Understanding the Problem

We need to determine the power of the lens that will allow a person with a near point of 30 cm to read a book placed at 50 cm. This is a lens problem that involves figuring out how much correction the lens needs to provide.
02

Using Lens Formula

The lens formula is given by \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \), where \( v \) is the image distance (here, it is the person's far point, 30 cm), \( u \) is the object distance (here, 50 cm), and \( f \) is the focal length of the lens. We need to find the focal length, \( f \).
03

Substitute Known Values

Substitute \( v = -30 \ \text{cm} \) (negative because image distance is taken as negative in this convention when on the same side as the light entering) and \( u = -50 \ \text{cm} \): \( \frac{1}{-30} - \frac{1}{-50} = \frac{1}{f} \).
04

Calculate Focal Length

Solve the equation:\[ \frac{1}{f} = \frac{1}{30} - \frac{1}{50} \]Find a common denominator and solve:\[ \frac{1}{f} = \frac{5 - 3}{150} = \frac{2}{150} = \frac{1}{75} \]Thus, \( f = 75 \ \text{cm} \).
05

Convert Focal Length to Power

The power of a lens in diopters (D) is given by the formula \( P = \frac{100}{f} \) when \( f \) is in centimeters. Substitute \( f = 75 \ \text{cm} \) to find:\[ P = \frac{100}{75} = 1.33 \] D.
06

Find the Required Negative Power

The given focal length is positive, but to correct the eyesight, a negative lens power is needed to bring the object closer. Therefore, the power required is \(-1.33 \ \text{D}\).

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Key Concepts

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

Lens Power Calculation
When it comes to correcting vision, calculating the lens power is essential. Lens power, measured in diopters (D), determines how converging or diverging the lens is. This power is inversely related to the focal length of the lens, measured in meters. To find the required power, we utilize the formula:
  • \( P = \frac{100}{f} \)
where \( f \) is the focal length in centimeters. For example, if the focal length is 75 cm, to find the power, simply divide 100 by 75, resulting in approximately 1.33 D. This process is crucial in determining how much the lens will refract light, essentially correcting the person's vision.
Lens Formula
The lens formula is a fundamental relationship that connects object distance (\( u \)), image distance (\( v \)), and focal length (\( f \)) of a lens. This relationship is given by:
  • \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
This equation helps us analyze and solve problems involving lenses, whether for magnification or correcting vision. For instance, in the given problem, \( v \) is the far point, and \( u \) is the distance of the book. By plugging these values into the formula, we find the focal length needed to correct the vision, which in turn helps calculate the appropriate lens power. This formula is pivotal for understanding how lenses bend light to form images at desirable locations.
Vision Correction
Many people require vision correction because their eyes cannot focus on objects at regular distances, a common problem addressed using lenses. For people who have trouble seeing clearly at a distance, like needing to read a book 50 cm away but only seeing clearly at 30 cm, lenses adjust the path of incoming light. They help the eyes to focus the light directly onto the retina.
Lenses needed for vision correction can be either
  • convex (for farsightedness)
  • or concave (for nearsightedness)
The problem presented is a case where a negative lens power is required, suggesting concave lenses. By making the lenses out of materials like plastic or glass, manufacturers customize the level of light bending required. Thus, understanding lens power and how to calculate it is key in creating effective solutions for those needing vision correction.

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Most popular questions from this chapter

The aperture of the largest telescope in the world is 5 \(\mathrm{m}\). If the separation between the moon and the earth is \(4 \times 10^{5} \mathrm{~km}\) and the wavelength of visible light is 5000 \(\AA\), then the minimum separation between the objects on the surface of the moon which can be just resolved is approximately (a) \(1 \mathrm{~m}\) (b) \(10 \mathrm{~m}\) (c) \(50 \mathrm{~m}\) (d) \(200 \mathrm{~m}\)

For observing a cricket match binoculars are preferred to a terrestrial telescope because (a) the binoculars give the proper three dimensions (b) the binoculars have a shorter length (c) the telescope does not give erect image (d) the telescope has chromatic aberration

To increase both the resolving power and magnifying power of a telescope (a) wavelength of light has to be decreased (b) aperture of the objective has to be increased (c) focal length of the objective has to be increased (d) both the focal length and aperture of the objective have to be increased

A concave spherical refracting surface with radius \(R\) separates a medium of refractive index \((5 / 2)\) from air. As an object is approaching the surface from far away from the surface along the central axis, its image (a) always remains real (b) always remains virtual (c) changes from real to virtual at a distance \((2 R / 3)\) from the surface (d) changes from virtual to real at a distance \((2 R / 3)\) from the surface

An object is approaching a fixed plane mirror with velocity \(5 \mathrm{~m} / \mathrm{s}\) making an angle of \(45^{\circ}\) with the normal. The speed of image with respect to mirror is (a) \(5 \mathrm{~m} / \mathrm{s}\) (b) \(\frac{5}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\) (c) \(5 \sqrt{2} \mathrm{~m} / \mathrm{s}\) (d) \(10 \mathrm{~m} / \mathrm{s}\)
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