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Chapter 34: Q47P (page 1040)

A double-convex lens is to be made of glass with an index of refraction of $1.5$ .One surface is to have twice the radius of curvature of the other and the focal length is to be $60 mm$ . What is the (a) smaller and (b) larger radius?

Short Answer

Expert verified

Smaller radius of the lens is $45 mm$
Larger radius of the lens is $90 mm$

Step by step solution

01

Listing the given quantities

Refractive index of lens, $n = 1.5$

Focal length, $f = 60.0 mm$

Radius of one surface is twice the other, i.e., $r_{2} = - 2 r_{1}$

02

Understanding the concepts of lens maker equation

By using the Lens maker鈥檚 formula, we can find the radius of both curvatures of the lens.

Formula:

Lens maker鈥檚 formula

$\frac{1}{f} = (n - 1) (\frac{1}{r_{1}} - \frac{1}{r_{2}}) . . . 34 - 10$

03

(a) Calculations of the smaller radius of the lens

Lens maker鈥檚 formula is given by equation,

$\frac{1}{f} = (n - 1) (\frac{1}{r_{1}} - \frac{1}{r_{2}})$

Since one surface has twice the radius of the other and since one surface is convex to the incoming light while the other is concave, we can write,

$r_{2} = - 2 r_{1}$

So, we can get,

$\frac{1}{f (n - 1)} = (\frac{1}{r_{1}} + \frac{1}{2 r_{1}}) = \frac{3}{2 r_{1}} r_{1} = \frac{3}{2} f (n - 1) = 1.5 (60.0) (1.5 - 1) = 45 mm$

Smaller radius of the lens is $45 mm$ .

04

(b) Calculations of the larger radius of the lens

As per the condition given in the problem, r²will be double of r₁

So, r₂ = 90 mm

Larger radius of the lens is 90 mm

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Most popular questions from this chapter

An object is placed against the center of a spherical mirror, and then moved $70 cm$ from it along the central axis as theimage distance $i$ is measured. Figure 34-36 gives $i$ versus object distance $p$ out to $p_{s} = 40 cm$ . What is $i$ for $p = 70 cm$ ?

17 through 29 22 23, 29 More mirrors. Object O stands on the central axis of a spherical or plane mirror. For this situation, each problem in Table 34-4 refers to (a) the type of mirror, (b) the focal distance $f$ , (c) the radius of curvature $r$ , (d) the object distance $p$ , (e) the imagedistance $i$ , and (f) the lateral magnification $m$ . (All distances are in centimeters.) It also refers to whether (g) the image is real $(R)$ or virtual localid="1662996882725" $(V)$ , (h) inverted $(I)$ or noninverted $(N I)$ from O, and (i) on the same side of the mirror as object O or on the opposite side. Fill in the missing information. Where only a sign is missing, answer with the sign.

Suppose the farthest distance a person can see without visual aid is $50 c m$ . (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 to $1 / f$ , where $f$ is in meters. What is $p$ for the lens?

In Fig. 34-32, an isotropic point source of light $S$ is positioned at distance $d$ from a viewing screen $A$ and the light intensity $I_{P}$ at point $P$ (level with $S$ ) is measured. Then a plane mirror $M$ is placed behind $S$ at distance $d$ . By how much is $I_{P}$ multiplied by the presence of the mirror?

You look down at a coin that lies at the bottom of a pool of liquid of depthand index of refraction(Fig. 34-57). Because you view with two eyes, which intercept different rays of light from the coin, you perceive the coin to bewhere extensions of the intercepted rays cross, at depth $d_{a}$ instead of $d$ . Assuming that the intercepted rays in Fig. 34-57 are close to a vertical axis through the coin, show that $d_{a} = \frac{d}{n}$ .

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