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Chapter 31: Problem 71

Show that placing a 1 -diopter lens in front of a 2 -diopter lens gives the equivalent of a single 3 -diopter lens (i.e., the powers of closely spaced lenses add).

Short Answer

Expert verified
Indeed, as per the law of combination of lenses, placing a 1 -diopter lens in front of a 2 -diopter lens gives the equivalent of a single 3 -diopter lens. The powers of closely placed lenses add up.

Step by step solution

01

Understanding lens power

The power of a lens is given by the reciprocal of its focal length. It is measured in diopters. If the focal length is measured in meters, then the power (P) in diopters is given by \(P = 1/f\). In this case, we are dealing with two individual lenses of 1 diopter and 2 diopters respectively.
02

Combining lens powers

According to the law of combination of lenses, the total power of the two lenses when they are placed together is the algebraic sum of their individual powers. Let's denote the power of the first lens as \(P1 = 1\) diopter and the power of the second lens as \(P2 = 2\) diopters. Then, the total power \(P_{total}\) would be \(P_{total} = P1 + P2 = 1 + 2 = 3\) diopters.

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Most popular questions from this chapter

A candle and a screen are \(70 \mathrm{cm}\) apart. Find two points between candle and screen where you could put a convex lens with \(17-\mathrm{cm}\) focal length to give a sharp image of the candle on the screen.

A lens has focal length \(f=35 \mathrm{cm} .\) Find the type and height of the image produced when a 2.2 -cm-high object is placed at distances (a) \(f+10 \mathrm{cm}\) and (b) \(f-10 \mathrm{cm} .\)

(a) Find the focal length of a concave mirror if an object placed \(38.4 \mathrm{cm}\) in front of the mirror has a real image \(55.7 \mathrm{cm}\) from the mirror. (b) Where and what type will the image be if the object is moved to a point \(16.0 \mathrm{cm}\) from the mirror?

A converging lens has focal length \(4.0 \mathrm{cm} .\) A 1.0 -cm-high arrow is located \(7.0 \mathrm{cm}\) from the lens with its lowest point \(5.0 \mathrm{mm}\) above the lens axis. Make a full-scale ray-tracing diagram to locate both ends of the image. Confirm using the lens equation.

How far from a \(25-\mathrm{cm}\) -focal-length lens should you place an object to get an upright image magnified 1.8 times?
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