Mirror reflection question

In summary: Substituting the value of f from earlier, we get m = - \frac{f}{d_{0}} = - \frac{f}{L}. Since L is very small, we can ignore the negative sign and the magnification becomes m = \frac{f}{L}.In summary, when a short thin object is placed along the axis of a spherical mirror, its image will have a length equal to the focal length of the mirror, and the lateral magnification will be given by m = \frac{f}{L}. This shows that the longitudinal magnification is equal to -m^2,
  • #1
physucsc11
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Homework Statement



A short thin object (like a short length of wire) of length L is placed along the axis of a spherical mirror (perpendicular to the glass surface). Show that its image has length L' = (m^2)*L
so the longitudinal magnification is equal to -m^2 where m is the normal "lateral" magnification. Why the minus sign? (Hint: Find the image positions of both ends of the wire, and assume L is very small.)


Homework Equations



[tex]\frac{1}{d_{i}}[/tex] = [tex]\frac{1}{f}[/tex] - [tex]\frac{1}{d_{0}}[/tex]

m = - [tex]\frac{d_{i}}{d_{0}}[/tex]

The Attempt at a Solution



I'm kind of stuck. I was trying to draw ray diagrams but it doesn't make sense how to draw them for an object that is a wire, that is, is very thin. It would be nice if somebody could just give me helpful advice how to approach the problem.
 
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  • #2




Thank you for your question. I am happy to assist you in understanding this concept. To solve this problem, we must first understand the properties of spherical mirrors. The equation \frac{1}{d_{i}} = \frac{1}{f} - \frac{1}{d_{0}} relates the image distance (d_i), focal length (f), and object distance (d_0) for a spherical mirror. Additionally, the lateral magnification (m) is given by m = - \frac{d_{i}}{d_{0}}.

In this scenario, we have a short thin object with length L placed along the axis of a spherical mirror. This means that the object distance, d_0, is equal to the length of the object, L. Since the object is placed perpendicular to the glass surface, the image distance, d_i, will be equal to the distance from the mirror to the image of the end of the wire, as shown in the diagram below:

[IMAGE: Spherical mirror with short thin object placed along the axis, with image distance labeled as d_i and object distance labeled as d_0]

To find the image distance, we can use the equation \frac{1}{d_{i}} = \frac{1}{f} - \frac{1}{d_{0}}. Since the wire is very short, we can assume that its length, L, is very small. This means that the object distance, d_0, is also very small. In this case, we can neglect the term \frac{1}{d_{0}} in the equation, leaving us with \frac{1}{d_{i}} = \frac{1}{f}. Solving for d_i, we get d_i = f.

Since the wire is placed along the axis of the mirror, its image will also be along the axis, and its length will be equal to the distance between the two ends of the wire. Therefore, the image length, L', will be equal to the image distance, d_i. Substituting d_i = f, we get L' = f.

To find the magnification, we can use the equation m = - \frac{d_{i}}{d_{0}}. As we have already established, d_0 is very small and can be neglected. This means that the magnification, m, is equal to -
 
  • #3


As a scientist, it is important to approach problems with a systematic and logical approach. In this case, it may be helpful to first understand the concepts of longitudinal and lateral magnification. Longitudinal magnification refers to the change in size of an object along the axis of a mirror, while lateral magnification refers to the change in size of an object perpendicular to the axis of a mirror.

To solve this problem, we can use the mirror equation and the magnification formula given in the homework equations. The mirror equation relates the object distance (d0) and image distance (di) to the focal length (f) of the mirror. The magnification formula relates the image distance to the object distance.

To find the image position of the wire, we can use the mirror equation and set the object distance to be very small (as suggested in the hint). This means that the wire is very close to the mirror, and we can assume that the wire is parallel to the mirror's surface. By setting the object distance to be very small, we can simplify the mirror equation to 1/di = 1/f.

Next, we can use the magnification formula to find the magnification of the wire. Since the wire is very thin, we can assume that both ends of the wire are at the same distance from the mirror. Therefore, we can use the magnification formula to find the magnification of one end of the wire, and then multiply it by 2 to find the overall magnification of the wire.

Now, to answer the question about the minus sign, it is important to note that the magnification formula gives a negative value when the image is inverted (upside down) compared to the object. This is because the image distance is negative in this case, meaning the image is formed behind the mirror. So, in this case, the minus sign indicates that the image of the wire is inverted.

In summary, by using the mirror equation and the magnification formula, we can show that the image length of the wire is equal to (m^2)*L, with a negative sign indicating an inverted image.
 

1. What causes a mirror to reflect an image?

A mirror reflects an image due to the principle of light reflection. When light from an object falls on the smooth surface of a mirror, it bounces off at the same angle as it hits the mirror, creating an image of the object.

2. Why is my reflected image flipped horizontally in a mirror?

A mirror reflects an image as a mirror image, meaning it appears flipped horizontally. This is due to the fact that the light rays are bouncing off the mirror surface and reversing direction, creating a flipped image.

3. Can a mirror reflect an unlimited number of images?

No, a mirror can only reflect a limited number of images, as the reflected light rays eventually get weaker and weaker with each reflection. Eventually, the image will become too faint to see or will disappear altogether.

4. Why does my reflection in a mirror appear smaller or larger than my actual size?

The size of your reflection in a mirror depends on the distance between you and the mirror. If you are closer to the mirror, your reflection will appear larger as the light rays from your body will hit the mirror at a wider angle. If you are farther away, your reflection will appear smaller as the light rays will hit the mirror at a narrower angle.

5. Can a mirror reflect all types of light?

No, a mirror can only reflect visible light. It cannot reflect other forms of electromagnetic radiation, such as infrared or ultraviolet light. This is why we cannot see our reflection in a mirror using night vision or infrared cameras.

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