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Chapter 33: Problem 17

Will the magnification produced by a simple magnifier increase, decrease, or stay the same when the object and the lens are both moved from air into water?

Short Answer

Expert verified
Answer: The magnification produced by a simple magnifier will decrease when the object and lens are both moved from air to water.

Step by step solution

01

Know the lens formula for air and water

The lens formula relates the focal length (f), object distance (u), and image distance (v) in a medium. The lens formula is given by: (1/f) = (n-1)((1/R1) - (1/R2)), where n is the refractive index of the lens with respect to the medium, R1 is the radius of curvature of the first surface, and R2 is the radius of curvature of the second surface. Since we are comparing the magnification when the lens and object are in air and water, we need to consider the lens formula in both media. In air, the refractive index is 1 while in water, it is greater than 1. We will use the refractive index (n) of the lens and water to find the new focal length in water.
02

Calculate the new focal length in water

To find the new focal length in water, we need to use the new refractive index (na = refractive index of lens in air, nw = refractive index of lens in water). The lens formula in water becomes: 1/fw = (na/nw - 1)((1/R1) - (1/R2)). Note that when the medium was air, the lens formula was: 1/fa = (na - 1)((1/R1) - (1/R2)). Divide the first equation by the second equation to get: fw/fa = (na - 1)/(na/nw - 1). Since na > 1 and nw > 1, fw > fa. The focal length in water is greater than the focal length in air.
03

Compare magnifications in air and water

We know that the magnification formula is M = (1 + D/f). Since the focal length in water (fw) is greater than the focal length in air (fa), the magnification in water will be less than the magnification in air. M_water = (1 + D/fw) < (1 + D/fa) = M_air. Therefore, when the object and lens are both moved from air to water, the magnification produced by a simple magnifier will decrease.

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Most popular questions from this chapter

Galileo discovered the moons of Jupiter in the fall of \(1609 .\) He used a telescope of his own design that had an objective lens with a focal length of \(f_{o}=40.0\) inches and an eyepiece lens with a focal length of \(f_{e}=2.00\) inches. Calculate the magnifying power of Galileo's telescope.

The radius of curvature for the outer part of the cornea is \(8.0 \mathrm{~mm}\), the inner portion is relatively flat. If the index of refraction of the cornea and the aqueous humor is 1.34: a) Find the power of the cornea. b) If the combination of the lens and the cornea has a power of \(50 .\) diopter, find the power of the lens (assume the two are touching).

An unknown lens forms an image of an object that is \(24 \mathrm{~cm}\) away from the lens, inverted, and a factor of 4 larger in size than the object. Where is the object located? a) \(6 \mathrm{~cm}\) from the lens on the same side of the lens b) \(6 \mathrm{~cm}\) from the lens on the other side of the lens c) \(96 \mathrm{~cm}\) from the lens on the same side of the lens d) \(96 \mathrm{~cm}\) from the lens on the other side of the lens e) No object could have formed this image.

LASIK surgery uses a laser to modify the a) curvature of the retina. b) index of refraction of the aqueous humor. c) curvature of the lens. d) curvature of the cornea.

Suppose the near point of your eye is \(2.0 \cdot 10^{1} \mathrm{~cm}\) and the far point is infinity. If you put on -0.20 diopter spec tacles, what will be the range over which you will be able to see objects distinctly?
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