### Video Transcript

A concave mirror of radius of curvature 10 centimeters is placed 30 centimeters to the right of a thin convex lens of focal length 15 centimeters. A small bulb sits 50 centimeters to the left of the lens. Find the distance to the left of the mirror of the image formed. Find the magnification produced by the system.

Letβs start by highlighting some of the vital information weβve been given in the statement. First, weβre told that the radius of curvature of a concave mirror is 10 centimeters. Weβll call that π. And weβre told that this mirror is located 30 centimeters to the right of a thin convex lens which has focal length 15 centimeters. Weβll call that π sub πΏ. We want to solve for two things. First, the distance to the left of the mirror that the imageβs formed which weβll call π sub ππ. And second, we want to solve for the magnification produced by the system of the mirror and the lens. Weβll call this π sub π .

Letβs start by drawing a diagram of this situation. In this situation, we have an optical system with two elements. Thereβs our thin convex lens. And then a concave mirror following the lens. And finally our object, a distance of 50 centimeters to the left of the lens. We want to solve first for the image distance formed by this system. Overall, what weβre going to do is solve for the image created by the lens which will look something like this. Weβre not exactly sure where or how tall it will be. But roughly, we know this is what it will look like. And then, this image formed by the lens becomes the object for the mirror. So the lens image is equal to the mirror object. Then weβll find the image that the mirror creates from this object. And the distance that image is from the mirror will be equal to π sub ππ.

So letβs start by solving for the image distance of the image that the lens creates, for now, completely ignoring the mirror. If our optical system consisted of only the object and the lens, then we would refer to the lens equation, sometimes also called the mirror equation, that says that the inverse of the focal length is equal to one over the object distance plus one over the image distance.

Now in our case as this first step, we want to solve for the image distance of the image created by the lens, where that cross in the middle of the lens on the optical axis is our reference point for the distances. If we algebraically rearrange this equation to solve for π sub π, we find that itβs equal to the focal length π times the object distance divided by the object distance minus the focal length. Now because we have an optical system with two components, the lens and the mirror, we want to be very careful to make sure weβre referring to the right thing when we write down a term. So letβs put some subscripts on these terms in this equation so we can identify what weβre talking about.

Weβre solving for the distance of the image created by the lens. So weβll call it π sub ππΏ. Thatβs equal to the focal length of the lens times the distance of the object from the lens divided by that object distance minus the focal length. Weβre told the focal length of the lens. And weβre also told that the object is a distance of 50 centimeters from the lens. Which means that π sub ππΏ is equal to 50 centimeters. When we plug in these two values and enter this fraction into our calculator, we find that the distance between the image formed by the lens and the lens is 21.43 centimeters.

Since the lens focal length is 15 centimeters, that is halfway between the lens and the mirror, this image distance tells us that the image created by the lens looks something like this. Itβs inverted and closer to the mirror than to the lens. Now remember that the image formed by the lens is equal to the object that the mirror uses to create an image.

Now letβs use the same equation to solve for image distance as before. But this time, instead of using the lens as a reference point, weβll use the mirror, abbreviated by π. So we want to solve for π sub π, the focal length of the mirror, and π sub ππ, the distance of the object from the mirror. To solve for π sub π, the mirrorβs focal length, we can recall the equation that relates focal length with radius of curvature π. The focal length of a lens or a mirror is equal to one-half its radius of curvature. In our case, the focal length of the mirror is π over two or 10 centimeters divided by two which is 5.0 centimeters.

Now to solve for π sub ππ, the distance of the object from the mirror β now to solve for π sub ππ, the distance of the object from the mirror, we can use the fact that the lens and mirror are separated by 30 centimeters. And that the distance between the lens and the image it forms is 21.43 centimeters. This means that π sub ππ equals 30 centimeters minus π sub ππΏ, or 21.43 centimeters. So we find that π sub ππ is 8.57 centimeters.

Weβre now ready to solve for π sub ππ, the image distance created by the mirror. When we calculate this fraction, we find that π sub ππ is equal to 12 centimeters. Thatβs how far to the left of the mirror the final image is formed.

Now that weβve solved for the final image distance, we want to know what is the overall magnification of this optical system of the lens plus mirror. To figure that out, letβs first recall some facts about magnification. The first thing we can remember, the first thing we can recall is that magnification π is equal to image height divided by object height which is also equal to negative image distance divided by object distance. And second, we recall that the total magnification for an optical system, which has π optical components, is equal to the magnification of the first component, π sub one, times the magnification of the second component, π sub two, and so on down to π sub π.

Now in our case, our system magnification π sub π is equal to the magnification of the lens times the magnification of the mirror. And we can use the fact that magnification equals negative image distance divided by object distance. When we plug in the image and object distances for the lens and mirror, we see that all four of these terms are now known. So letβs plug in for each one. When we enter these four values into our calculator, we find that the overall system magnification is positive 0.60. This means that, compared to the size of our object, our final image is six 10ths as big.