(II) A 4.2-cm-tall object is placed 26 cm in front of a spheri...

Question

(II) A 4.2-cm-tall object is placed 26 cm in front of a spherical mirror. It is desired to produce a virtual image that is upright and 3.2 cm tall.

(a) What type of mirror should be used?

(b) Where is the image located?

(c) What is the focal length of the mirror?

(d) What is the radius of curvature of the mirror?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with mirrors. So this problem says in a small toy store, a customer is trying to create a fun display for kids using a toy car. The toy car has a height of 3.8 centimeters and is positioned 25 centimeters away from a spherical mirror. The customer wants to achieve an erect virtual image of the car that measures three centimeters in height. There are four parts to this question. Part one. What type of mirror would the customer need to produce such an erect virtual image? For part two, where, where will this new image of the toy car form relative to the mirror? For part three, what is the focal length of the mirror required for this scenario? And for part four, what is the radius of curvature of this mirror? We were given four possible choices as our answers for choice. A four point or part one, the type of mirror co is convex part two, the image distance is negative 20 centimeters. For part three, the focal length is negative 94 centimeters. And for part four, the radius of curvature is a negative 1.9 multiplied by 10 to the two centimeters. For choice B for part one, the type of mirror is convex part two, the image distance is 20 centimeters. Part three, the focal length is 94 centimeters. And from part four, the radius of curvature is 1.9 multiplied by 10 to the two centimeters. For choice C for part one, the type of mirror is concave. For part two, the image distance is negative 20 centimeters. For part three, the focal length is negative 94 centimeters. And for part four, the radius of curvature is negative 1.9 multiplied by 10 to the two centimeters. And for choice D for part one, the type mirror is concave. Part two. The image distance is 20 centimeters. Part three, the focal length is 94 centimeters. And for part four, the radius of curvature is 1.9 multiplied by 10 to the two centimeters. They were off by looking at part one. And here we need to identify the type of mirror that we need whether a con cave or a convex mirror. Now, we're told that we need an erect virtual image and both convex and concave mirrors are capable of producing erect virtual images. But we're also given the heights of the object and the image. So the object height is 3.8 centimeters and the image height is going to be three centimeters. So that means our image is actually going to be smaller than our object and only a convex mirror is capable of producing such an image. So with virtual images for a concave mirror, the image is always going to be larger than the object. So in this case, we're gonna need a convex mirror based on the image heights. So that's gonna be our answer to part one is convex. Now, for part two, we need to figure out um where the image is going to be formed. And for this, we're actually going to use our magnification to recall that our magnification M is equal to R image height H I divided by our object height ho but this magnification is also equal to minus D I divided by do where D I here is our image distance and do is our object distance. So we're gonna want to solve this for D I, we'll have D I is equal to minus H I divided by ho and that fraction is multiplied by do. Now, we, we already have looked at our problem and we've identified our image height and our object height. But we were also given the object distance, which is the 25 centimeters. So we actually have values for everything on the right hand side. So we can plug those in. And so we'll have D I, which is our image distance. It's going to be equal to minus our image height, which is H I was three centimeters that gets divided by the object height, which is the 3.8 centimeters. And that fraction is multiplied by the object distance, which was 25 centimeters. So when I plug those values into my calculator, I get D I is equal to minus 19.74 centimeters. And I'm gonna need to round that to two significant figures. And I went round that to just two significant figures. I yet D I is equal to negative 20 centimeters. So that's gonna be our answer for part two. Now, for part three, I need to find the focal length. And for this, we're going to use our mirror equation. So recall that one divided by do plus one divided by D I is equal to one divided by F. And weve just calculated our image distance and we were given the object distance so we can plug those values in and then solve for F. So I'll have for the object distance, I'll have the 25 centimeters. So I'll have one divided by 25 centimeters plus one divided by. And here we're gonna use our um image distance that we just calculated, which is gonna be a negative 19.74 centimeters. And that's gonna be equal to one divided by F. And so you just take the inverse to both sides. So I'll have F is equal to the quantity of one divided by 25 centimeters minus one divided by 19.74 centimeters and that quantity is raised to the negative one power. And so when I calculate F, I'll get negative 93.8 centimeters. And again, I'm gonna have to round that to two significant figures. And so that means my F is going to be equal to negative 94 centimeters says come may answer to part three. And then finally, we need to calculate the radius of curvature. And for that, we're going to need to recall our relationship between the focal length and the radius of curvature. So recall that F is going to be equal to R divided by two, where R here is gonna be a radius of curvature and F is the focal length. So solved this for R and so I'll just multiply over the two. So I have R as equal to two F and we just calculated the focal length. So we can plug that in. So we'll have R is going to be equal to two multiplying the negative 93.8 centimeters. And so R is gonna be equal to and here, I'll just go ahead and round it to two significant figures of negative 1.9 multiplied by into the two centimeters. So this can be our answer for part four. So now that I have uh answers for all four parts, I can look at my answer choices and this actually corresponds to answer choice. A so just a quick little recap of what we've done here, we started off by determining the type of mirror. And here we had to use the fact that we had an erect virtual image. But we also had to use the fact that the object uh height was actually greater than the image height, which cannot occur for a concave mirror. So that meant that our mirror had to be a convex mirror. For part two, we used our magnification in both forms in terms of the height and the distances. And we use that to calculate the image distance. For part three, we used our mirror equation along with our object distance and our newly calculated image distance to calculate the focal length. And then finally, for part four, we just use a relationship between the radius of curvature and the focal length to calculate the radius of curvature. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Types of Mirrors

Mirror Formula

Magnification

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