An object is 16.0 cm to the left of a lens. The lens forms an 36.0 cm to the right of the lens. (c) Draw a principal-ray diagram.

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A bug is placed 10.0 centimeters in front of a lens. The lens forms an image on a white screen, 30.0 centimeters behind the lens, draw a ray diagram to show the formation of the image. So that is our end goal is to draw a ray diagram to show the formation of the image that the bug makes. OK. So we're given some multiple choice answers here. So we have four different ray diagrams to choose from. And only one of them is the correct answer. The gray little oval right here represents the lens. We also have a blue arrow that represents the bug. And then on the left side of the lens represents the bug. And then on the right side of the lens will represent the image. And the only one that shows it's not on the left side for the bug image is for, for multiple choice answer B OK. And we show our different rays for F one and F two for each of the multiple choice answers A through D OK. So to solve this problem, we need to recall and use the equation for the object image relationship for a lens. And that states that one divided by S plus one, divided by S prime is equal to one divided by F and S is the object distance which in this case, it's 10.0 centimeters and S prime is the image distance which in this case is 30.0 centimeters and F represents the focal length. So now we need to solve for the focal length since that is our unknown variable. So we need to rearrange and solve for F. So using a little bit of algebra and rearrange the sulfur F or focal length. So we get that F is equal to S multiplied by S prime divided by S plus S prime. So now we can plug in all of our known values in sulfur F. So S is 10.0 centimeters multiplied by 30.0 centimeters divided by 10.0 centimeters plus 30.0 centimeters. So when we plug it into a calculator, we will determine that our focal length is 7.5 centimeters. So since the focal length is greater than zero, that means that it's converging. So the lens I should say is converging to be clear. OK. So now we must recall and use the lateral magnification equation. And the lateral magnification is represented by the letter M. So M is equal to S prime divided by S which we can plug in our known variable. So S prime is 30.0 centimeters divided by 10 point zero centimeters. But since it's converging, this is negative. So when we plug that into a calculator, we should get minus three as our answer. So now let us draw a diagram to help us support the information we just determined. So I went ahead and drawn a complex diagram to help us understand what's going on here. So ray one, which as we can see right here, I'll highlight it in yellow really quickly is here. And ray one is drawn parallel to the principal axis. After being refracted by the lens, this ray passes through the focal point F two on the back side of the lens. OK. And ray two which I've highlighted in yellow again is drawn through the center of the lens and continues in a straight line as shown which I'm gonna highlight it again in pink. So you can see here. So that is ray too. And ray three is drawn through the focal point F one on the front side of the lens and emerges from the lens parallel to the principal axis. So I'm gonna highlight that in green. So this is ray three awesome. So therefore the final image is real and is inverted. So that means we'll write our correct answer in green. The correct answer has to be the letter A. So let's go look at our multiple choice answers to see which of these ray diagrams looks like the one we just drew. So using our handy dandy green highlighter, the correct answer is a thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.