The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the of the arrow formed by paraxial rays incident on the convex surface. Is the erect or inverted?

Video transcript

Welcome back, everyone. We are told that a lapidary cuts and polishes one side of a cylinder shaped diamond into a spherical surface. And we are told that the spherical surface has a radius of six centimeters. Now, we are also told that the N value for our diamond is going to be 2.41. Now, what we do is we place a Rubik's cube with a height of 10 centimeters and we place it at a position that is 30 centimeters to the left of the vertex. Now, we are tasked with finding two things. One, what is going to be the final position of our image and two, what is going to be the height of our image of the Rubik's cube? Now, for reference here, one more variable we're gonna need to know uh we are told that the end value for air is simply one. Now, before getting started here, I do wish to acknowledge our multiple choice answers here on the left hand side of the screen, those are going to be the values that we strive for. So without further ado let us begin. All right. Well, where do we begin? Well, for a spherical surface, we have the following relation here that the end value for our first material in this case, er divided by the position of our object plus the end value for our second material. In this case, the diamond divided by the position of our image is going to be equal to NB minus N A divided by the radius of our spherical surface. I'm gonna go ahead and change colors here and we'll go ahead and get started on part one. I'm gonna take the above equation and I'm going to subtract N A over. So on both sides of our equation now what this leaves me is NB divided by SI is equal to and B minus N A divided by R minus N A over. So finally to isolate my SI variable on the left hand side, I'm just going to divide both sides by NB. You'll see on the left hand side, the NB cancel out on top and bottom and let's go ahead and then fill in the values that we know. What we get is one over the position of our image is equal to 2.41 minus one, divided by six minus 1/ all divided by 2.41. What this gives us is that one over si is equal to 1/1, 1.95 taking the reciprocal of both sides. What we get is that the position of our image is equal to 11.95 centimeters. All right, let me go ahead and change colors here and we'll move on to part two. Now, in order to find the height of our image, we are going to need to know what our lateral magnification is. Now, once again, for a spherical surface, we actually have an equation for this. This is going to be the negative of N A times si divided by and B times. So what this is going to be equal to is negative one times 11. divided by 2. times 30. Now, what is this going to give us? Well, this is going to give us negative 0.165. And since our lateral magnification is negative, this means that we are going to have a real image. Now we are ready to calculate our image height. And here is the equation. The magnitude of our image height is going to be the absolute value of our lateral magnification times the absolute value or the magnitude of our object height. Now what this gives us is negative 0.165 times the absolute value of 10 giving us 1.65 centimeters. So now we have found both the final position of our image as well as the height of our image. And both of these answer choices correspond to our final answer. Choice of a. Thank you all so much for watching. I hope this video helped. We will see you all in the next one.