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Chapter 31: Problem 31

A tiny insect is trapped 1.0 mm from the center of a spherical dewdrop \(4.0 \mathrm{mm}\) in diameter. As you look straight into the drop, what's the insect's apparent distance from the drop's surface?

Short Answer

Expert verified
The insect's apparent distance from the dewdrop surface is \(1.25 \, \text{mm}\).

Step by step solution

01

Identify the Given Elements

From the problem, we know the actual distance of the insect from the center of the dewdrop \(d = 1.0\) mm and the radius of the dewdrop \(r = 2.0\) mm (half of the diameter). Also, the refractive index of water (\(n_{water}\)) is typically 1.333.
02

Use the Law of Refraction (Snell's Law)

According to Snell's law, the apparent distance (depth) is given by the actual depth divided by the refractive index of the medium. So the apparent depth \(d'\) is given by \(d' = d / n_{water}\).
03

Calculate the Apparent Distance

Substituting the values, we get the apparent distance \(d' = 1.0 \, \text{mm} / 1.333 \approx 0.75 \, \text{mm}\). This is the apparent distance of the insect from the center of the dewdrop.
04

Calculate Apparent Distance from the Surface

The problem, however, asks for the apparent distance from the drop’s surface. So, subtract the apparent distance to the center from the radius: \( \text{Apparent Distance from Surface = Radius - Apparent Distance to Center} = 2.0 \, \text{mm} - 0.75 \, \text{mm} = 1.25 \, \text{mm}\).

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Most popular questions from this chapter

An object and its lens-produced real image are \(2.4 \mathrm{m}\) apart. If the lens has 55 -cm focal length, what are the possible values for the object distance and magnification?

For visible wavelengths, the refractive index of a thin glass lens is \(n=n_{0}-b \lambda,\) where \(n_{0}=1.546\) and \(b=4.47 \times 10^{-5} \mathrm{nm}^{-1} .\) If its focal length is \(30 \mathrm{cm}\) at \(550 \mathrm{nm}\), how much does the focal length vary over a wavelength spread of 10 nm centered on 550 nm?

You're an optician who's been asked to design a new replacement lens for cataract patients. The lens must be 5.5 mm in diameter, with focal length \(17 \mathrm{mm}\), and it can't be thicker than \(0.8 \mathrm{mm} .\) For the lens material, you have a choice of plastic with refractive index 1.49 or more expensive silicone with \(n=1.58 .\) Which material do you choose, and why?

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A 12 -mm-high object is \(10 \mathrm{cm}\) from a concave mirror with focal length \(17 \mathrm{cm} .\) (a) Where is the image, (b) how high is it, and (c) what type is it?
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