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Chapter 34: Problem 16

A tank whose bottom is a mirror is filled with water to a depth of 20.0 \(\mathrm{cm} .\) A small fish floats motionless 7.0 \(\mathrm{cm}\) under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?

Short Answer

Expert verified
(a) 5.26 cm, (b) 20.30 cm

Step by step solution

01

Understanding the Concept

When an object is viewed through a different medium, such as water, it appears at a different position than its actual position due to refraction. The apparent depth is given by the formula \( d' = \frac{d}{n} \), where \( d \) is the actual depth and \( n \) is the refractive index of water (approximately 1.33).
02

Calculating Apparent Depth of the Fish

The fish is actually 7 cm below the water surface. Using the formula \( d' = \frac{d}{n} \), substitute \( d = 7 \) cm and \( n = 1.33 \). The apparent depth \( d' = \frac{7}{1.33} \approx 5.26 \) cm.
03

Understanding the Image in the Mirror

The bottom of the tank is a mirror, so an image of the fish will be formed at an additional depth equal to the distance from the fish to the mirror. The mirror creates a virtual image at the same distance below it as the original fish above.
04

Calculating Path to Image in the Mirror

The total distance from the water surface to the image in the mirror is the sum of the depth of the fish (7 cm) and the distance from the mirror to the water surface (20 cm). This is 27 cm.
05

Calculating Apparent Depth of Mirror Image

The apparent depth of the image is calculated by using the entire 27 cm distance as the actual depth, leading to \( d'' = \frac{27}{1.33} \approx 20.30 \) cm.

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Key Concepts

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

Refraction
When light passes through different materials, its speed changes. This bending of light is known as refraction. If you're looking at something underwater, like a fish, refraction makes it seem like it's in a different spot than it really is. Imagine bending a pencil by placing part of it in a glass of water, it looks broken at the water's surface, due to the light bending.
  • Refraction causes objects to look as if they are at a shallower depth than they actually are.
  • This happens because light travels slower in water compared to air, altering its path.
Understanding refraction is key to explaining why a fish looks closer to the water surface than it really is.
Refractive Index
The refractive index is a number that tells us how much light slows down in a material. Each substance has its own refractive index. For example, the refractive index of water is around 1.33.
  • This means light travels 1.33 times slower in water than in a vacuum.
  • The apparent depth can be calculated using the formula: \( d' = \frac{d}{n} \), where \( d \) is the actual depth and \( n \) is the refractive index.
For the fish at 7 cm depth, using this formula gives us an apparent depth of approximately 5.26 cm when viewed from above, showing how much closer the fish seems to be.
Virtual Image
A virtual image is formed when light rays appear to meet at a location, but do not actually converge there. In our scenario with the fish and the mirror at the tank's bottom, the virtual image looks real to the observer, even though no fish is present at that location in the water.
  • The mirror at the bottom of the tank creates an image of the fish as if below the mirror.
  • The fish's apparent image is then deeper than the actual mirrored image due to refraction.
By calculating the apparent depth of the fish and mirror image, we find that the fish appears to be at a depth of approximately 20.30 cm, considering the combined effects of the mirror and refraction.

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Most popular questions from this chapter

Resolution of a Microscope. The image formed by a microscope objective with a focal length of 5.00 \(\mathrm{mm}\) is 160 \(\mathrm{mm}\) from its second focal point. The eyepiece has a focal length of 26.0 \(\mathrm{mm}\) . (a) What is the angular magnification of the microscope? (b) The unaided eye can distinguish two points at its nicroscope? (b) The unaided eye can distinguish two points at its near point as separate if they are about 0.10 \(\mathrm{mm}\) apart. What is the minimum separation between two points that can be observed (or resolved) through this microscope?

Figure 34.41 shows photographs of the same scene taken with the same camera with lenses of different focal length. If the object is 200 \(\mathrm{m}\) from the lens, what is the magnitude of the lateral magnification for a lens of focal length (a) \(28 \mathrm{mm} ;\) (b) 105 \(\mathrm{mm}\) ; (c) 300 \(\mathrm{mm}\) ?

A double-convex thin lens has surfaces with equal radii of curvature of magnitude 2.50 \(\mathrm{cm} .\) Looking through this lens, you observe that it forms an image of a very distant tree at a distance of 1.87 \(\mathrm{cm}\) from the lens. What is the index of refraction of the lens?

A telescope is constructed from two lenses with focal lengths of 95.0 \(\mathrm{cm}\) and \(15.0 \mathrm{cm},\) the 95.0 -cm lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall, 3.00 \(\mathrm{km}\) away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

Contact Lenses. Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 \(\mathrm{cm}\) from his eyes instead of the usual 25.0 \(\mathrm{cm} .(\mathrm{a})\) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?
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