/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 A drop of water on a countertop ... [FREE SOLUTION] | 91Ó°ÊÓ

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A drop of water on a countertop reflects light from a flower held \(3.0 \mathrm{cm}\) directly above it. The flower's diameter is \(2.0 \mathrm{cm},\) and the diameter of the flower's image is \(0.10 \mathrm{cm} .\) What is the focal length of the water drop, assuming that it may be treated as a convex spherical mirror?

Short Answer

Expert verified
The focal length of the water drop is approximately 0.143 cm.

Step by step solution

01

Recognize the Known Values

We know the object distance \(d_o = 3.0\, \text{cm}\), the object size \(h_o = 2.0\, \text{cm}\), and the image size \(h_i = 0.10\, \text{cm}\). We can use these to find the image distance \(d_i\) using magnification.
02

Use the Magnification Formula

Magnification \(M\) is given by the formula \(M = \frac{h_i}{h_o} = \frac{d_i}{d_o}\). Substituting the known values gives \(\frac{0.10}{2.0} = \frac{d_i}{3.0}\).
03

Solve for Image Distance

To find \(d_i\), solve \(\frac{0.10}{2.0} = \frac{d_i}{3.0}\), which gives \(d_i = \frac{0.10}{2.0} \times 3.0 = 0.15\, \text{cm}\).
04

Use the Mirror Equation

In spherical mirrors, the mirror equation is given by \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). Substitute \(d_o = 3.0\, \text{cm}\) and \(d_i = 0.15\, \text{cm}\) into this equation.
05

Calculate the Focal Length

Plug the values into the mirror equation: \(\frac{1}{f} = \frac{1}{3.0} + \frac{1}{0.15}\). This simplifies to \(\frac{1}{f} = \frac{1}{3.0} + \frac{1}{0.15}\), which evaluates to \(\frac{1}{f} = 0.333 + 6.667 = 7.000\), so \(f = \frac{1}{7.000} \approx 0.143\, \text{cm}\).

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Key Concepts

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

Focal Length
The focal length is a key concept when dealing with spherical mirrors, like the one described in the exercise. It is the distance from the mirror's surface to the focal point, where light rays that are parallel to the principal axis converge (for concave mirrors) or appear to diverge from (for convex mirrors). In the context of a convex spherical mirror, as in our water drop example, the focal length is considered as a negative value because the focal point is virtual, meaning light rays do not actually converge; instead, the parallel rays appear to diverge as if they were coming from a focal point behind the mirror. To find the focal length, you can use the mirror equation, which is a valuable tool in calculating how light behaves with spherical mirrors. Remember, the focal length provides insight into how powerful the mirror is at converging or diverging light.
Magnification Formula
Magnification is about how much larger or smaller the image is compared to the actual object. For a convex spherical mirror, this involves understanding how image size and object size relate to one another through the magnification formula. The magnification (M) is defined as:\[M = \frac{h_i}{h_o} = \frac{d_i}{d_o}\]Where:- \(h_i\) is the height of the image- \(h_o\) is the height of the object- \(d_i\) is the distance from the image to the mirror- \(d_o\) is the distance from the object to the mirrorIn the exercise, you learned how to calculate the magnification and use it to find other unknown variables like the image distance. By solving the formula, you can determine how much the flower's reflection in the water drop is reduced in size, giving insight into the nature of the spherical mirror at play. Especially for a convex mirror, the magnification is always less than 1, indicating a reduced image size.
Mirror Equation
The mirror equation is fundamental in optics, providing a relationship between the object distance, the image distance, and the focal length of a spherical mirror. The equation is expressed as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]This means:- \(f\) is the focal length- \(d_o\) is the object distance- \(d_i\) is the image distanceIn our example with the water drop, you use known values to plug into this equation and find unknowns like the focal length. It shows how altering object distance affects the focal length or vice versa. This equation helps in understanding how real or virtual images form and where they appear when light rays interact with mirrors. It’s crucial for mastering concepts like image formation, which depend heavily on the physics embodied by the mirror equation. By understanding this interplay, you deepen your grasp on how images form on curved surfaces.

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Most popular questions from this chapter

A small statue has a height of \(3.5 \mathrm{cm}\) and is placed in front of a concave mirror. The image of the statue is inverted, \(1.5 \mathrm{cm}\) tall, and located \(13 \mathrm{cm}\) in front of the mirror. Find the focal length of the mirror.

A spacecraft is in a circular orbit about the moon, \(1.22 \times 10^{5} \mathrm{m}\) above its surface. The speed of the spacecraft is \(1620 \mathrm{m} / \mathrm{s}\), and the radius of the moon is \(1.74 \times 10^{6} \mathrm{m} .\) If the moon were a smooth, reflective sphere, (a) how far below the moon's surface would the image of the spacecraft appear, and (b) what would be the apparent speed of the spacecraft's image? (Hint: Both the spacecraft and its image have the same angular speed about the center of the moon.)

A dentist's mirror is placed \(2.0 \mathrm{cm}\) from a tooth. The enlarged image is located \(5.6 \mathrm{cm}\) behind the mirror. (a) What kind of mirror (plane, concave, or convex) is being used? (b) Determine the focal length of the mirror. (c) What is the magnification? (d) How is the image oriented relative to the object?

An object is located \(14.0 \mathrm{cm}\) in front of a convex mirror, the image being \(7.00 \mathrm{cm}\) behind the mirror. A second object, twice as tall as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height as the other image. How far in front of the mirror is the second object located?

A concave mirror \((f=45 \mathrm{cm})\) produces an image whose distance from the mirror is one-third the object distance. Determine (a) the object distance and (b) the (positive) image distance.

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