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Chapter 20: Problem 104

The power and type of the lens by which a person can see clearly the distant objects, if a person cannot see objects beyond \(40 \mathrm{~cm}\), are : (a) \(-2.5 \mathrm{D}\) and concave lens (b) \(-2.5 \mathrm{D}\) and convex lens (c) \(-3.5 \mathrm{D}\) and concave lens (d) \(-3.5 \mathrm{D}\) and convex lens

Short Answer

Expert verified
The power is \(-2.5 \mathrm{D}\) and the lens is concave; option (a) is correct.

Step by step solution

01

Understanding the Problem

A person who cannot see objects beyond 40 cm is likely nearsighted, or myopic. We need to determine the power and type of lens required for this person to clearly see distant objects.
02

Formula for Lens Power

The lens formula is given by \( \, \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \, \). For distant vision correction, the far point \( u \) is at infinity, making the formula \( \, \frac{1}{f} = \frac{1}{v} \, \). Here, \( v \) is -40 cm (negative as it is virtual) and we need the focal length \( f \).
03

Calculate the Focal Length

Using the modified formula: \( \frac{1}{f} = \frac{1}{-0.4} \). Converting 40 cm to meters gives us \( \frac{1}{-0.4} \approx -2.5 \) m. Therefore, \( f = -0.4 \) m.
04

Determining Lens Power

The power \( P \) of the lens is calculated as \( P = \frac{1}{f} \) where \( f \) is in meters. So, \( P = \frac{1}{-0.4} = -2.5 \) D.
05

Identifying Lens Type

Since the power is negative, the lens type must be concave. Concave lenses are used for correcting myopia (nearsightedness).
06

Select the Correct Option

From the options given, the correct one is (a) \(-2.5 \mathrm{D}\) and concave lens.

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Key Concepts

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

Myopia Correction
Myopia, or nearsightedness, is a common vision condition where close objects appear clearly, but distant ones are blurry. This is usually due to the eyeball being too long or the cornea having excessive curvature, which causes light rays to focus in front of the retina instead of on it.

To correct myopia, the use of lenses is essential.
  • The lenses help shift the focus back onto the retina.
  • This is accomplished by diverging light rays slightly before they enter the eye.
Corrective lenses for myopia are always concave because they spread out the incoming light rays, making it easier for the eye to focus them on the retina. This correction allows individuals with myopia to see distant objects clearly.
Concave Lens
A concave lens, also known as a diverging lens, is characterized by its inward-curving surfaces. These lenses are thinner at the center than at the edges. The primary function of a concave lens is to diverge light rays passing through it.
  • This divergence makes the rays appear to originate from a point behind the lens.
  • This is why they create a virtual image of objects, allowing viewers to see them as if they were further away.
Concave lenses are crucial in correcting vision defects like myopia. They help shift the focal point of light rays to fall directly on the retina.
In practical applications, concave lenses are used in eyeglasses, camera lenses, and telescopes, supporting functions that require image correction or enlargement.
Focal Length Determination
Determining the focal length of a lens is an essential process in understanding how lenses work to correct myopia. The focal length, represented by the letter \( f \), is the distance between the lens and the point where light rays converge or appear to diverge from a single point.

To find the focal length of a corrective lens:
  • Use the lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \).
  • For distant objects, assume the far point \( u \) to be at infinity. This simplifies the formula to \( \frac{1}{f} = \frac{1}{v} \).
In the case of myopia correction, pilots or virtual images position \( v \)鈥攆or a person who cannot see beyond 40 cm鈥攏eeds to be calculated.
When \( v \) is -40 cm (converted to -0.4 m), the focal length \( f \) becomes -0.4 m, and thus the lens power \( P \) is \( -2.5 \) D.
Understanding these calculations helps patients and opticians determine the required lens specifications for effective vision correction.

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Most popular questions from this chapter

If a crown glass prism of refracting angle \(10^{\circ}\) have refractive indices for red and violet rays \(1.514\) and \(1.523\) respectively, then the dispersion caused by a crown glass prism is : (a) \(0.07^{\circ}\) (b) \(0.08^{\circ}\) (c) \(0.09^{\circ}\) (d) \(0.10^{\circ}\)

The electric permittivity and magnetic permeability of free space are \(\varepsilon_{0}\) and \(\mu_{0}\), respectively. The index of refraction of the medium, if \(\varepsilon\) and \(\mu\) are the electric permittivity and magnetic permeability in a medium is : (a) \(\frac{\varepsilon \mu}{\varepsilon_{0} \mu_{0}}\) (b) \(\left(\frac{\varepsilon \mu}{\varepsilon_{0} \mu_{0}}\right)^{1 / 2}\) (c) \(\frac{\varepsilon_{0} \mu_{0}}{\varepsilon \mu}\) (d) \(\left(\frac{\varepsilon_{0} \mu_{0}}{\varepsilon \mu}\right)^{1 / 2}\)

A planet is observed by an astronomical refracting telescope having an objective of focal length \(16 \mathrm{~cm}\) and eye-piece of focal length \(20 \mathrm{~cm}\). Then : (a) the distance between objective and eye-piece is \(16.02 \mathrm{~m}\) (b) the angular magnification of the planet is 800 (c) the image of the planet is inverted (d) both (a) and (b) are correct

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In the figure, a point source \(' P^{\prime}\) is placed at a height \(h\) above the plane mirror in a medium of refractive index \(\mu\). An observer \(O\), vertically above \(P\), outside the liquid, sees \(P\) and its image in the mirror. The apparent distance between these two is: (a) \(2 \mu h\) (b) \(\frac{2 h}{\mu}\) (c) \(\frac{2 h}{\mu-1}\) (d) \(h\left(1+\frac{1}{\mu}\right)\)
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