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Chapter 34: Q140P (page 1046)

Suppose the farthest distance a person can see without visual aid is50cm. (a) What is the focal length of the corrective lens that will allow the person to see very far away? (b) Is the lens converging or diverging? (c) The power Pof a lens (in diopters) is equal to1/f, wherefis in meters. What ispfor the lens?

Short Answer

Expert verified
  1. Focal length of the corrective lensf=-0.50cm
  2. The lens is diverging
  3. Power of the lensP=-2.0diopters

Step by step solution

01

Listing the given quantities

Far point of the personi=50cm

02

Understanding the concepts of lens formula and focal length

We will use the lens formula to find the focal length of the lens. The reciprocal of the focal length gives the power of the lens. If the focal length of the lens is negative, it is a diverging lens

Formula:

1f=1p+1i

Power of the lens,

role="math" localid="1663046890549" 1f=p

03

Calculations of the focal length of the corrective lens

(a)

Let’s take the object distance to be at infinity, i.e.,p=∞.

The image is inverted, therefore,theni=-0.50m

1f=1p+1i=1∞+1-0.50

Hence, the focal length of the lens isf=-0.50m

04

Type of the lens

(b)

As f<0, therefore the lens is diverging.

05

Calculations of the power of the lens

(c)

The power of the lens is defined as

P=1f=1-0.50=-2.0diopters

Power of the lensrole="math" localid="1663047184088" P=-2.0diopters

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Most popular questions from this chapter

Figure 34-47a shows the basic structure of the human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens (Fig. 34-47b). A “normal” eye can focus parallel light rays from a distant object O to a point on the retina at the back of the eye, where the processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina (Fig. 34-47c). (a) Suppose that for the parallel rays of Figs. 34-47a and b, the focal length f’of the effective thin lens of the eye is 2.50 cm. For an object at distance p = 40 cm, what focal length f’ of the effective lens is required for the object to be seen clearly? (b) Must the eye muscles increase or decrease the radii of curvature of the eye lens to give focal length f’?

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