Can someone help me with this hw problem

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In summary: So, the image in the convex mirror is 1/2 the size of the image in the plane mirror. That means the distance between the mirror and the image in the convex mirror is twice the distance between the mirror and the image in the plane mirror. In summary, a child standing between two mirrors, one being a plane mirror and the other a convex mirror, will see their image in the plane mirror appear to be twice as tall as their image in the convex mirror. This is due to the actual height of the image in space and its distance from the mirror. The child states that the angle subtended by the image in the plane mirror is twice the angle subtended by the image in the convex mirror. To
  • #1
rossi02
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A child stands at the midpoint between two mirrors, which are separated by 10.0 m. One of the mirrors is a plane mirror, and the other mirrors is convex. The child looks at the two images and notices that her image in the plane mirror appears to be exactly twice as tall as her image in the convex mirror. This appearance is from her viewpoint - it does not represent the actuasl height of the images. The appearracne to the child will be based on two contributions - the actual height of the image in space and how far away the image is . Thus, acccoding to the child, the angle subtended by the image in the plane mirror is twice the angle subtended by the image in the convex mirror. what is the focal length of the convex mirror.

thanks
 
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  • #2
The "actual" height of the image of the child in the plane mirror is the height of the child. The image appears to be 20 feet away (the mirror is ten feet away and the image appears to be 10 feet behind the mirror).

Let "f" be the focal length of the convex mirror and set up the formulas for size and distance to the image based on that. The apparent size of an image depends directly on the actual size of the image and inversely on the distance to the image (that's from "similar triangles"!).
 
  • #3
for reaching out for help with this problem! It seems like the key to solving this problem is understanding how mirrors work and how they create images. In this case, the child is standing at the midpoint between the two mirrors, so the distance from the child to each mirror is equal. This means that the images in both mirrors will appear to be the same distance away from the child.

The key difference between the two images is their size. The plane mirror creates a virtual image that is the same size as the object, while the convex mirror creates a virtual image that is smaller than the object. This is because convex mirrors have a negative focal length, meaning that the focal point is behind the mirror.

To find the focal length of the convex mirror, we can use the relationship between the object distance (distance from the object to the mirror), the image distance (distance from the image to the mirror), and the focal length. This relationship is known as the mirror equation: 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.

Since the child is standing at the midpoint between the two mirrors, the object distance for both mirrors will be the same (5m). We also know that the image in the plane mirror appears to be twice the size of the image in the convex mirror. This means that the image distance for the plane mirror (di1) will be twice the image distance for the convex mirror (di2).

Using this information, we can set up the following equations:
1/f = 1/5 + 1/di1
1/f = 1/5 + 1/(2*di2)

Solving for di2 in the second equation, we get di2 = 10/3 m.

Now, using the first equation and plugging in di2, we can solve for the focal length:
1/f = 1/5 + 1/(10/3)
1/f = 3/10
f = 10/3 m

Therefore, the focal length of the convex mirror is 10/3 m. I hope this helps you understand and solve the problem. Good luck!
 

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