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An object is placed \(10.0 \mathrm{~cm}\) from a convex mirror having a focal length of magnitude \(15.0 \mathrm{~cm} .\) (a) Use graph paper to construct an accurate ray diagram for this situation. From your ray diagram, determine the location of the image, its magnification, and whether it is upright or inverted. (b) Check your answers to part (a) using the mirror equation.

Short Answer

Expert verified
By drawing a ray diagram, we determined that the virtual image formed by the convex mirror is upright and smaller than the object with a magnification of less than 1. Verification through mirror equation confirmed the accuracy of image location and its measurements.

Step by step solution

01

Drawing a Ray Diagram

To begin, use graph paper to draw an accurate ray diagram. First, place your mirror (the line on the graph). Then situate the object 10 cm from the mirror. Next, draw at least two rays: One parallel to the principle axis (this reflects back along the line from the focal point), and one through the focal point (this reflects back parallel). If the rays diverge, extend them behind the mirror to locate the image.
02

Determine Image Location and Magnification

The point where the extended rays intersect on the back side of the mirror is the image location. Measure this distance from the mirror - this gives the image distance. As this is a virtual image created by a convex mirror, the magnification will be less than 1. Determine the height of the image and divide it by the height of the object to get the magnification.
03

Identify Nature of Image

Now we can observe if the image is upright or inverted by comparing its orientation with the object. As it is formed by a convex mirror, it should be upright.
04

Verify using the Mirror Equation

Let's validate these findings by using the mirror equation 1/f = 1/v + 1/u. Here, f is the focal length, v is the image distance and u is the object distance. Remember to consider the sign convention: For a virtual image, v should be negative. Solve this equation for v, which should match the image distance obtained in Step 2.

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Key Concepts

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

Ray Diagram
In the context of convex mirrors, constructing a ray diagram is a vital tool in understanding how images are formed. To draw a ray diagram accurately, start by representing the convex mirror on graph paper. Position the object at a fixed point, say 10 cm from the mirror's surface. From the top of this object, draw the first ray parallel to the principal axis. Naturally, this ray reflects and appears to originate from the focal point behind the mirror.

The second ray should be drawn as if it passes through the focal point and reflects back parallel to the principal axis. For convex mirrors, these reflected rays will diverge. Hence, you must extend these lines backward to find their apparent intersection point on the opposite side of the mirror. This intersection represents the virtual image. Remember, it's important to use a ruler for precision and to draw neatly so that the results are as accurate as possible.

Ray diagrams assist not only in visually confirming the presence of a virtual image but also in estimating where this image appears to form in relation to the mirror.
Mirror Equation
The mirror equation, \[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\] is crucial for confirming the details obtained in the ray diagram. In this equation:
  • \(f\) is the focal length of the mirror.
  • \(v\) is the image distance.
  • \(u\) is the object distance from the mirror.
Always apply the sign convention: for convex mirrors, both the focal length \(f\) and image distance \(v\) would be negative.

Insert the known values: \(f = -15 \text{ cm}\) and \(u = 10 \text{ cm}\) into the equation. Solve for \(v\), to find the image distance. This result should align with the intersection point found in the ray diagram analysis.

The mirror equation provides a mathematical method to verify where and how an image is formed, thus adding a layer of precision to the interpretations.
Virtual Image
Convex mirrors always form virtual images. This means that the images you see cannot be projected onto a screen, as they form at a point that the light rays appear to diverge from, rather than actually converging. Virtual images are characterized by the fact that they are located at a position where rays appear to intersect, but do not physically do so.

In a convex mirror scenario, these images appear smaller and upright relative to the object. This happens because the light rays are reflected outward from the mirror's surface, but since they never actually converge, the image is virtual.

Understanding the nature of virtual images in convex mirrors helps in practical applications like car side mirrors, where a wider field of view and upright images are desired.
Magnification
Magnification describes how much larger or smaller the image is compared to the object. In the context of convex mirrors, magnification is always less than one, reflecting the fact that the image appears smaller than the object.

The formula for magnification \(M\) is given by:\[M = \frac{h'}{h} = \frac{v}{u}\]where:
  • \(h'\) is the height of the image.
  • \(h\) is the height of the object.
  • \(v\) is the image distance.
  • \(u\) is the object distance.
The negative sign in the formula indicates that the image formed is virtual, as it appears upright in a convex mirror. Calculate \(M\) using the measurements from your ray diagram or mirror equation. This value tells us exactly how much smaller the image is when compared to the original object size.

By examining magnification, you gain insight into the size relationships between objects and their images, furthering your understanding of convex mirrors' applications.

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Most popular questions from this chapter

Object \(\mathrm{O}_{1}\) is \(15.0 \mathrm{~cm}\) to the left of a converging lens with a \(10.0-\mathrm{cm}\) focal length. A second lens is positioned \(10.0 \mathrm{~cm}\) to the right of the first lens and is observed to form a final image at the position of the original object \(\mathrm{O}_{1}\). (a) What is the focal length of the second lens? (b) What is the overall magnification of this system? (c) What is the nature (i.e., real or virtual, upright or inverted) of the final image?

A jellyfish is floating in a water-filled aquarium \(1.00 \mathrm{~m}\) behind a flat pane of glass \(6.00 \mathrm{~cm}\) thick and having an index of refraction of \(1.50 .\) (a) Where is the image of the jellyfish located? (b) Repeat the problem when the glass is so thin that its thickness can be neglected. (c) How does the thickness of the glass affect the answer to part (a)?

One end of a long glass rod \((n=1.50)\) has the shape of a convex surface of radius \(8.00 \mathrm{~cm} .\) An object is positioned in air along the axis of the rod in front of the convex surface. Find the image position that corresponds to each of the following object positions: (a) \(20.0 \mathrm{~cm}\), (b) \(8.00 \mathrm{~cm}\), (c) \(4.00 \mathrm{~cm}\), (d) \(2.00 \mathrm{~cm}\).

The lens-maker's equation for a lens with index \(n_{1}\) immersed in a medium with index \(n_{2}\) takes the form $$ \frac{1}{f}=\left(\frac{n_{1}}{n_{2}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) $$ A thin diverging glass (index \(=1.50\) ) lens with \(R_{1}=\) \(-3.00 \mathrm{~m}\) and \(R_{2}=-6.00 \mathrm{~m}\) is surrounded by air. An arrow is placed \(10.0 \mathrm{~m}\) to the left of the lens. (a) Determine the position of the image. Repeat part (a) with the arrow and lens immersed in (b) water (index \(=1.33\) ) (c) a medium with an index of refraction of \(2.00\). (d) How can a lens that is diverging in air be changed into a converging lens?

A contact lens is made of plastic with an index of refraction of \(1.50 .\) The lens has an outer radius of curvature of \(+2.00 \mathrm{~cm}\) and an inner radius of curvature of \(+2.50 \mathrm{~cm}\). What is the focal length of the lens?

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