a-concavemirror-produces-a-virtual-image-that-is-three-times-as-tall-as-the-object-aifthe-object-is-

Question

A concavemirror produces a virtual image that is three times as tall as the object. (a)Ifthe object is 16 $\mathrm{cm}$ in front of the mirror, what is the
image distance? (b) What is the focal length of this mirror?

David Kobylarz · Accepted Answer

Okay, So for this question, we know that in a conclave mirror, when the objects will be on the center of curvature, a real image is going to be formed that is basically reduced in size and inverted relative to the object. And when the object this placed between the focal point of the center of curvature or real images formed, that would be in Lord and inverted relative to the object. And when the object is located between the con Cape mirror and its focal point in large, upright and virtual images produced. So for port A, we have a concave mirror produces a virtual image that is three times as tall as the object. So the magnification of the mirrors three the information that is produced by the mirror we know by the following equation em He was the distance image led by distance of the object negative. And we here we would have to solve for the d I the distance of the object. Therefore we rearrange and we get that the distance with the object equals the Magnifico negative magnification distance of the on here. We substance to what we have in, so we know that we know that the magnification is three. Look in three and that the distance of the object is 16 centimeters. And therefore, when we do this, you know the distance of the image is negative. 48 centimeters and then for part B, we are going to find the focal length of this mirror, so we would go ahead. And for part B, you would use the following equation. One over the focal length equals one over the distance of the object Los one over the distance of the image. And here we would just plug in what we know. But we can really combined these terms and therefore have that one over F. He calls distance of the image close distance of the object, divided by distance of the inimitable deployed by the distance of the object. And then, since we have the inversion, what the focal length we would have, The focal length is equal to distance of the image times the distance of the object, divided by distance of the image, plus the distance of the object. And now we would go ahead and plug in all the values that we know, So we would plug in the F equals life equals 16 centimeters multiplied by the negative 48 wintry, calculated before divided by a negative 48 plus 16. Not only do this mass, we get that the focal, like of the conclave mirror in question is 24 centimeters.

David Kobylarz · Answer

Okay, so for this question, port A. We will be looking for to the image distance with the given information. And we know at the con Cave mirror would produce a real image that is three times as tall as the object. So we know that the height of the image equals three times the height of the object, then magnificent. Therefore, we can also say that the magnification would equal height of the image divided by height of the object. And then we would substitute the negative three for em since, well, that&#x27;s be the magnification here. So we would have negative three equals negative distance of the image about by distance of the object. Therefore, distance of the image one equal three times the distance of the object. And here we can substitute. Since we know the distance of the object, we can substitute in the 22 centimeters which is given so that the distance of the image equals three times a 22 centimeters and therefore we get the distance of the image is equivalent to 66 centimeters. Because of this sense of the image from the mirror and now for poor be, we will once again calculates the focal length of the mirror. Using the Formula One over the focal length equals one over the distance of the object, this one divided by the distance of the image. And since we know that the distance of the object this twe two centimeters from the distance of the image, it&#x27;s 66 centimeters. We just plugged that in, so we have that one over. The focal length was one divided by 22 centimeters, plus one over 66 centimeters, and this is also equivalent to 4/66 centimeters, and therefore local ING is equivalent to 16.62 centimeters, which is approximately 17 centimeters for the focal length.

David Kobylarz · Answer

Okay, so this first part of this question, we&#x27;re looking for the distance of the image from the mirror and we know that the magnification of the mirror is 1/3. So here we would have used we would use the equation. Magnification equals negative distance of the image divided by the distance of the object. You can, in theory, move this old equation around to find the distance of the image which would be equal to the negative magnification times the distance of the object. And since we know the distance of the object from a mere two B 27 centimeters and the magnification to be 1/3 who we just plugged that into this formula to find the distance of the image. So you have negative times. Magnification shows 1/3 multiplied by the distance of the object from the mirror, which is 27 centimeters. Which means that the distance of the image is negative. 27 divided by three. Therefore, the distance of the image is negative. Mine centimeters from the mirror. And similarly, what does kitten With this information, we can also calculate the focal length of the mirror by using the equation. One over F equals one over. I didn&#x27;t know the assistance of the object, just one over the distance of the image, and here we can plug it in the distance of the object. Therefore, one over equals the distance of the object, which is 27 centimeters plus one over the negative nine, which is the distance of the image. And if we calculate everything and take the inverse of everything, we can see that we get a Focal Inc that equals negative 13.5 seven years.

Mayukh Banik · Answer

In this particular case, it is said that the image is located to help centimeter behind the meter, which means that the images march jewel immense distance sees negative 12.12 centimeter. We know that magnification is equal to evens distance over object distance. And in this particular case, magnification is say to be to third. So using these, we can have the object distance. It was 3/2 times the image distance We just 12 and we can their plus minus over here. We&#x27;ll figure out the sign later. So this times out to be 18 centimeter. We, uh, last minus sign. Now we will use the thence form of the F equals one over b O. Thus one over b i in verse of that object distance that is just food. Both the values the last minus 18 minus one of what has putting the sign off the lunch with image. So this comes out to be putting all negative signs. Using negative 18 we will get using negative 18. We get negative seven point do centimeter Using the positive sign, we get positive. 3.6 centimeter positive means this is Ah Khan gave meter So this is out of question. So focus on and off convex meter that will do that job will be negative 7.2 centimeters.

Mayukh Banik · Answer

the object is located. The image is located at negative 36 centimeter. Because the image is behind the meter, that is what is a virtual image and we have to use a negative sign. Corresponding. Do the image distance magnification is given by half, which is image distance over object distance. So object distance seekers to ice the image. Distance equals 72 centimeter. Now we can use the meter. For me, the one over F equals one over object distance. That&#x27;s wonderful. Emails. Distance equals one of a 72 minus one of what, 36. This comes out to be negative one of a 72. So if equals negative 72 centimeter is the desire focus lint off the convicts made.

Mayukh Banik · Answer

the first set of information that is given these object distance equals 38 0.4 centimeter image distance is 55.7 centimeters. So the for President of the Lens is given by Essen Board Plus s prime in verse. It wasn&#x27;t the whole thing that he&#x27;s 38.4 in verse. The US 55.7 in divorce in verse of the entire it was we will do the math now 38.4 in verse. Thus 55.7 in horse. It wasn&#x27;t that comes out to be ready to burn 73 So I&#x27;m going to take one decimal place, 22.7 centimeter in the next part. They are asking s prime equals question Mark when the images distance is 16 sandy meter. So you do the math again. One of our F equals one of what s thus one would s prime from theirs. If you saw, you will have s prime or s. It was one over if minus one over s prime. In the words of that which is 22.7 in horse minus 16 in verse. In the worse It was negative. 54.2 centimeter. There&#x27;s a correction. We&#x27;re trying to find ESP rhyme. That&#x27;s this is a spry in on this is s over here. Okay, so finally, virtual image 54.2 centimeter away.

Problem

A concave mirror produces a real image that is th…

Problem 36 Easy Difficulty

A concavemirror produces a virtual image that is three times as tall as the object. (a)Ifthe object is 16 $\mathrm{cm}$ in front of the mirror, what is the
image distance? (b) What is the focal length of this mirror?

Answer

(a) Thus, the image distance is $[-48 \mathrm{cm}]$
(b) Thus, the focal length of the concave mirror is $[24 \mathrm{cm}]$

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James S. Walker

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Liev B.

Marshall S.

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Wave Optics - Intro

Multiple Aperture Interference - Overview

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(a) Thus, the image distance is $[-48 \mathrm{cm}]$(b) Thus, the focal length of the concave mirror is $[24 \mathrm{cm}]$

Physics

Christina K.

Liev B.

Marshall S.

Farnaz M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Christina K.

Liev B.

Marshall S.

Farnaz M.

(a) Thus, the image distance is $[-48 \mathrm{cm}]$
(b) Thus, the focal length of the concave mirror is $[24 \mathrm{cm}]$

James S. Walker

Physics