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Chapter 34: Problem 18

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
The apparent depth of the fish when viewed at normal incidence is 5.26 cm. The depth of the image of the fish when viewed at normal incidence is 14 cm.

Step by step solution

01

Calculate Apparent Depth of the Fish

Apparent depth is the depth that an object seems to be at due to the refraction of light in a transparent medium. When light leaves a medium of high refractive index (water, n=1.33) and enters a medium of lower refractive index (air, n=1.00), it speeds up and bends away from the normal. The formula to find apparent depth is \( \text{apparent depth} = \text{real depth} \times \frac{\text{refractive index of air}}{\text{refractive index of water}} \). Using the provided data, it's \( \text{apparent depth} = 7.0 cm \times \frac{1.00}{1.33} = 5.26 cm \).
02

Calculate Image Depth of the Fish

The fact that there's a mirror at the bottom of the tank creates a reflection of the fish, creating an image. This image depth comes from two components: the actual depth of the fish and the apparent depth of the fish's image in the mirror. Hence, to find the image depth seen by the observer: \( \text{image depth} = \text{actual depth} + \text{apparent depth from the mirror} \). From step 1, the apparent depth from the mirror will be the apparent depth of the fish above the mirror, which will be the same as the actual depth of the fish from the water's surface (due to symmetry). So, the total image depth will be \( \text{image depth} = 7 cm + 7 cm = 14 cm \).

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Key Concepts

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

Refraction of Light
When light travels through different mediums, it can bend in a phenomenon known as refraction. This bending occurs due to the change in speed of light as it moves from one medium to another. Imagine dipping a straight stick halfway into water; the stick appears bent at the water's surface. This occurs because light is refracted at the boundary between air and water.

Refraction is responsible for a range of everyday observations, including the apparent depth of objects under water. When looking at a fish in a tank, the fish seems closer to the surface than it truly is. This is because the light rays coming from the fish bend away from the normal (an imaginary line perpendicular to the surface) as they move from water to air, making the fish appear at a shallower depth.
Refractive Index
The refractive index (or index of refraction) of a medium is a number that describes how much light bends when entering the medium. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. A higher refractive index means that light travels more slowly in the medium and the bending will be more pronounced.

For instance, water has a refractive index of around 1.33, while air's refractive index is approximately 1.00. When light exits from water (with a high refractive index) to air (with a lower refractive index), it speeds up and bends away from the normal. To calculate the apparent depth of an object in water, we use the formula: \[ \text{apparent depth} = \text{real depth} \times \frac{\text{refractive index of air}}{\text{refractive index of water}} \]. Using this formula, we can understand why objects under water look closer to the surface than they really are.
Mirror Reflection
Mirror reflection is quite different from refraction; here, light bounces off a surface rather than bending through it. The law of reflection states that the angle of incidence (the angle between the incoming ray and the normal) is equal to the angle of reflection (the angle between the reflected ray and the normal).

This principle leads to the formation of images in mirrors: if you place an object in front of a mirror, the light reflecting off the object will bounce off the mirror surface at the same angle, forming an image that seems to exist behind the mirror. For an observer, this results in the phenomenon where the image's depth appears to be the sum of the distance between the object and the mirror, and the distance from the mirror to the surface. Hence, for a fish in a tank with a mirrored bottom, its reflected image appears deeper than the fish itself due to the combination of its real depth and the light's journey after reflection.

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Most popular questions from this chapter

For a concave spherical mirror that has focal length \(f=+18.0 \mathrm{~cm},\) what is the distance of an object from the mirror's vertex if the image is real and has the same height as the object?

You wish to project the image of a slide on a screen \(9.00 \mathrm{~m}\) from the lens of a slide projector. (a) If the slide is placed \(15.0 \mathrm{~cm}\) from the lens, what focal length lens is required? (b) If the dimensions of the picture on a \(35 \mathrm{~mm}\) color slide are \(24 \mathrm{~mm} \times 36 \mathrm{~mm},\) what is the minimum size of the projector screen required to accommodate the image?

(a) You want to use a lens with a focal length of \(35.0 \mathrm{~cm}\) to produce a real image of an object, with the height of the image twice the height of the object. What kind of lens do you need, and where should the object be placed? (b) Suppose you want a virtual image of the same object, with the same magnification-what kind of lens do you need, and where should the object be placed?

A small tropical fish is at the center of a water-filled, spherical fish bowl \(28.0 \mathrm{~cm}\) in diameter. (a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?

The left end of a long glass rod \(8.00 \mathrm{~cm}\) in diameter, with an index of refraction of 1.60 , is ground and polished to a convex hemispherical surface with a radius of \(4.00 \mathrm{~cm}\). An object in the form of an arrow \(1.50 \mathrm{~mm}\) tall, at right angles to the axis of the rod, is located on the axis \(24.0 \mathrm{~cm}\) to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?
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