So the sheet of paper should be placed 24 cm from the mirror.

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The discussion focuses on the placement of a sheet of paper for projecting an image using a concave mirror with a focal length of 16 cm. The correct formula to use is 1/f = 1/p + 1/q, where p represents the object distance and q represents the image distance. By establishing that the image distance (q) is twice the object distance (p), the calculations reveal that the sheet of paper should be placed 24 cm from the mirror. This is derived from solving the equation and confirming that p equals 8 cm, leading to the final placement distance.

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Mirror due Midnight!

Problem 32.
A concave spherical mirror can be used to project an image onto a sheet of paper, allowing the magnified image of an ulluminated real object to be accurately traced.
If you have a concave mirror with a focal length of 16 cm, where would you place a sheet of paper so that the image projected onto it is twice as far from the mirror as the object is? In units of cm.
Note: I figured out that the formula I should use is 1/f=1/p+1/q
so would it be 1/f=2/-q+1/q
If so i need help to go on if not i need more help please this is due by Midnight.
f=16
 
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mustang said:
Problem 32.
A concave spherical mirror can be used to project an image onto a sheet of paper, allowing the magnified image of an ulluminated real object to be accurately traced.
If you have a concave mirror with a focal length of 16 cm, where would you place a sheet of paper so that the image projected onto it is twice as far from the mirror as the object is? In units of cm.
Note: I figured out that the formula I should use is 1/f=1/p+1/q
so would it be 1/f=2/-q+1/q
If so i need help to go on if not i need more help please this is due by Midnight.
f=16
You didn't define q and p, but I think you're trying to say that p = 2q (image distance equals twice the object distance). Why put the 2 in the numerator, then? If p = 2q:

[tex]\frac{1}{f} = \frac{1}{2q} + \frac{1}{q}[/tex]

[tex]\frac{2}{f} = \frac{1}{q} + \frac{2}{q}[/tex]

[tex]\frac{2}{f} = \frac{3}{q}[/tex]

[tex]q = \frac{(3)(f)}{2}= \frac{48}{2} = 24 cm[/tex]
 
cm, p=object distance, q=image distance

In order to find the distance at which the sheet of paper should be placed, we can use the formula 1/f = 1/p + 1/q. Since we know that the focal length (f) is 16 cm, we can rearrange the formula to solve for q: 1/q = 1/f - 1/p. Now, we also know that the image distance (q) should be twice the object distance (p), so we can substitute 2p for q in the formula: 1/(2p) = 1/16 - 1/p. We can then solve for p by cross-multiplying: 2p = 16p - 16. Simplifying, we get p = 8 cm. Therefore, the sheet of paper should be placed 24 cm (16 cm focal length + 8 cm object distance) from the mirror in order for the image to be twice as far from the mirror as the object is. Remember to always double check your calculations and units to ensure accuracy. Good luck with your assignment!
 

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