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Chapter 26: Problem 14

Amber \((n=1.546)\) is a transparent brown-yellow fossil resin. An insect, trapped and preserved within the amber, appears to be \(2.5 \mathrm{cm}\) beneath the surface when viewed directly from above. How far below the surface is the insect actually located?

Short Answer

Expert verified
The actual depth of the insect is 3.865 cm below the surface.

Step by step solution

01

Understanding Refraction

Light bends when it passes from one medium to another due to refraction. When viewing the insect through amber, light travels from amber into air, causing it to bend back and make the insect appear closer to the surface than it actually is.
02

Refraction Formula

We'll use the formula for apparent depth due to refraction: \( d_a = \frac{d_r}{n} \), where \( d_a \) is the apparent depth, \( d_r \) is the real depth, and \( n \) is the refractive index of amber. Rearrange it to find the real depth: \( d_r = d_a \times n \).
03

Substituting Known Values

The apparent depth \( d_a = 2.5 \) cm and the refractive index \( n = 1.546 \). Substitute these values into the rearranged equation: \( d_r = 2.5 \times 1.546 \).
04

Calculating Real Depth

Perform the multiplication to find \( d_r \): \( d_r = 2.5 \times 1.546 = 3.865 \) cm. This is the actual distance the insect is from the surface of the amber.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refractive Index
The refractive index is a measure of how much a ray of light bends when it enters a different medium from air. Imagine it as the unique bending fingerprint of each material. This bending occurs because light waves slow down or speed up as they encounter materials of different densities. Every transparent substance, like water, glass, or amber, has its own refractive index.
For instance:
  • If a material has a high refractive index, light bends more sharply upon entering it.
  • A lower refractive index means less bending of light.
The formula used for refractive index is:\[ n = \frac{c}{v} \]where \(c\) is the speed of light in a vacuum and \(v\) is the speed of light in the medium. In our example, amber has a refractive index of 1.546, making it bend light in a way that alters our perception of depth.
Apparent Depth
When you look through a medium like amber, the object beneath seems closer to you than it truly is. This is known as 'apparent depth'. The bending of light rays causes this effect.
To find the apparent depth, you use:\[ d_a = \frac{d_r}{n}\]where \(d_a\) is the apparent depth, \(d_r\) is the real depth, and \(n\) is the refractive index. Notice how light appears to travel less by making the object look shallower. It is the visual trickery of refraction!This concept is vital in understanding not only natural observations like an insect in amber but also in practical applications such as underwater photography or designing lenses. The insect in amber appears to be 2.5 cm below the surface due to the refractive index of amber which alters our perception.
Real Depth
The real depth is the actual measurement of how deep an object is beneath a surface in a medium. It differs from apparent depth due to the bending of light, which tricks our eyes into seeing an object in a slightly different position.
Calculating real depth is important because it gives us the true distances despite what our eyes perceive. To find the real depth, we rearrange the refraction equation:\[ d_r = d_a \times n\]Given the apparent depth (like the 2.5 cm for the insect in amber) and the refractive index of the medium (here, 1.546), calculating:\[ d_r = 2.5 \times 1.546 = 3.865 \, \text{cm}\]shows us that the insect is actually 3.865 cm below the surface. Understanding real depth helps in scientific studies and practical situations, ensuring accuracy beyond what is visually perceived.

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Most popular questions from this chapter

The drawing shows a ray of light traveling through three materials whose surfaces are parallel to each other. The refracted rays (but not the reflected rays) are shown as the light passes through each material. A ray of light strikes the \(a-b\) interface at a \(50.0^{\circ}\) angle of incidence. The index of refraction of material \(a\) is \(n_{a}=1.20 .\) The angles of refraction in materials \(b\) and \(c\) are, respectively, \(45.0^{\circ}\) and \(56.7^{\circ} .\) Find the indices of refraction in these two media.

Visitors at a science museum are invited to sit in a chair to the right of a full-length diverging lens \(\left(f_{1}=-3.00 \mathrm{m}\right)\) and observe a friend sitting in a second chair, \(2.00 \mathrm{m}\) to the left of the lens. The visitor then presses a button and a converging lens \(\left(f_{2}=+4.00 \mathrm{m}\right)\) rises from the floor to a position \(1.60 \mathrm{m}\) to the right of the diverging lens, allowing the visitor to view the friend through both lenses at once. Find (a) the magnification of the friend when viewed through the diverging lens only and (b) the overall magnification of the friend when viewed through both lenses. Be sure to include the algebraic signs \((+\) or \(-\) ) with your answers.

The near point of a naked eye is \(32 \mathrm{cm} .\) When an object is placed at the near point and viewed by the naked eye, it has an angular size of 0.060 rad. A magnifying glass has a focal length of \(16 \mathrm{cm},\) and is held next to the eye. The enlarged image that is seen is located \(64 \mathrm{cm}\) from the magnifying glass. Determine the angular size of the image.

A slide projector has a converging lens whose focal length is \(105.00 \mathrm{mm}\) (a) How far (in meters) from the lens must the screen be \(10 \mathrm{c}-\) ated if a slide is placed \(108.00 \mathrm{mm}\) from the lens? (b) If the slide measures \(24.0 \mathrm{mm} \times 36.0 \mathrm{mm},\) what are the dimensions (in \(\mathrm{mm}\) ) of its image?

Bill is farsighted and has a near point located \(125 \mathrm{cm}\) from his eyes. Anne is also farsighted, but her near point is \(75.0 \mathrm{cm}\) from her eyes. Both have glasses that correct their vision to a normal near point \((25.0 \mathrm{cm}\) from the eyes), and both wear the glasses \(2.0 \mathrm{cm}\) from the eyes. Relative to the eyes, what is the closest object that can be seen clearly (a) by Anne when she wears Bill's glasses and (b) by Bill when he wears Anne's glasses?
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