One end of a long glass rod (n=1.5) is a convex surface of radius 6.0 cm. An object is located

One end of a long glass rod (n=1.5) is a convex surface of...

One end of a long glass rod $(n=1.5)$ is a convex surface of radius $6.0 \mathrm{~cm}$. An object is located in air along the axis of the rod, at a distance of $10 \mathrm{~cm}$ from the convex end. (a) How far apart are the object and the image formed by the glass rod? (b) Within what range of distances from the end of the rod must the object be located in order to produce a virtual image?

One end of a long glass rod $(n=1.5)$ is a convex surface of radius $6.0 \mathrm{~cm}$. An object is located in air along the axis of the rod, at a distance of $10 \mathrm{~cm}$ from the convex end. (a) How far apart are the
object and the image formed by the glass rod? (b) Within what range of distances from the end of the rod must the object be located in order to produce a virtual image?

Key Concepts

Refraction at Spherical Surfaces

This concept describes how light bends when passing through a curved interface between two media with different refractive indices. The relationship among the object distance, image distance, refractive indices, and the radius of curvature is described by the spherical refraction formula (n?/s + n?/s' = (n? - n?)/R). This formula is fundamental in predicting image location and characteristics when light transitions between media at a spherical boundary.

Sign Conventions in Geometrical Optics

In geometrical optics, assigning the correct sign to distances and radii is essential for correct application of formulas. These conventions determine whether distances are considered positive or negative based on the direction of light propagation and the curvature of optical surfaces, playing a crucial role in distinguishing between real and virtual images.

Virtual Image Formation

A virtual image is formed when light rays appear to diverge from a point behind an optical element, even though they do not actually converge there. The concept is important because virtual images cannot be captured on a screen, and their formation is typically associated with a negative image distance in the chosen sign convention, indicating that the image is located on the same side as the object.

Transcript

00:01 Given the information for the index of refraction of the first medium, n sub 1, the index every fraction of the second medium, n sub 2, and we know n sub 1 because it's an error, and air has index every fraction of 1.

00:11 The radius of curvature of the surface r, as well as the distance p between the object and the lens here, we want to figure out for part a, how far apart are the object and the image formed by the glass rod? so first we need to find the distance to the image, i, and then we can figure out the difference between p and i so we can use the lend maker equation n1 over p plus n2 over i is equal to the difference between the two n2 minus n1 over r so we need to solve for i right so we'll just start doing that so we find that n2 over i is equal to n2 minus n1 over r minus minus minus n1 over p.

01:15 Or in other words, i is equal to n2 divided by everything that was on the right side of that equation.

01:25 So this would be n2 minus n1.

01:29 Let's write that a little neater.

01:31 Minus n1 over r minus n1 over p.

01:40 So we know everything on the right side of the equation.

01:42 We just simply plug them in, and we find a value for i of negative 90 centimeters.

01:46 So that negative sign means that the image is going to be on the same side as the object.

01:52 So then to find the distance between the image and the object, we're just going to call that d.

01:56 It's going to be the absolute value of i because they're on the same side.

02:00 So we're going to get rid of that negative sign minus p.

02:04 So it'll be 80 centimeters or excuse me 90 centimeters minus 80 centimeters, which comes, or excuse me, 90 centimeters minus 10 centimeters, which comes out to be equal to 80 centimeters.

02:18 And we can box that in as our solution for part a.

02:24 Part b has asked to find the range of distances from the end of the rod, the object has to be located in order to produce a virtual image...

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