A Glass Rod. Both ends of a glass rod with index of...

A Glass Rod. Both ends of a glass rod with index of refraction 1.60 are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is $6.00 \mathrm{cm},$ and the radius of curvature at the right end is 12.0 $\mathrm{cm} .$ The length of the rod between vertices is 40.0 $\mathrm{cm} .$ The object for the surface at the left end is an arrow that lies 23.0 $\mathrm{cm}$ to the left of the vertex of this surface. The arrow is 1.50 $\mathrm{mm}$ tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (d) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image?

A Glass Rod. Both ends of a glass rod with index of refraction 1.60 are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is $6.00 \mathrm{cm},$ and the
radius of curvature at the right end is 12.0 $\mathrm{cm} .$ The length of the rod between vertices is 40.0 $\mathrm{cm} .$ The object for the surface at the left end is an arrow that lies 23.0 $\mathrm{cm}$ to the left of the vertex of this surface. The arrow is 1.50 $\mathrm{mm}$ tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (d) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image?

Key Concepts

Magnification

Magnification describes how the size (and sometimes the orientation) of an image compares to that of the original object. It is determined by the ratio of the image height to the object height and is influenced by the geometry of the system. In multi-surface optical systems, the overall magnification is the product of the magnifications from each refracting or reflecting element, offering insight into the final image’s size and orientation.

Spherical Refracting Surface

A spherical refracting surface is a boundary between two media where at least one side of the boundary is part of a sphere. In geometric optics, such surfaces cause incident light rays to refract following Snell’s law. The curvature of the surface (expressed by its radius of curvature) determines how the rays converge or diverge, thus playing a critical role in image formation.

Refraction Equation for Spherical Surfaces

The refraction at a spherical interface is often analyzed using the spherical refraction formula, which relates the object distance, image distance, refractive indices of the two media, and the radius of curvature of the surface. This equation, usually written as (n1/object distance) + (n2/image distance) = (n2 - n1)/radius of curvature, is essential for determining the position and nature of images formed by curved surfaces.

Sign Conventions in Geometric Optics

Proper analysis of optical systems requires a consistent set of sign conventions to assign positive or negative values to distances, radii, and heights. These conventions help determine whether an image or object is real or virtual, and they are crucial when applying formulas to interfaces, especially in systems with multiple elements. The sign convention dictates the treatment of distances measured in the direction of light propagation versus those measured against it.

Real and Virtual Images

Real images are formed where light rays actually converge and can be projected onto a screen, while virtual images occur when diverging rays appear to originate from a point, so they cannot be directly captured. Identifying whether an image or object is real or virtual is key in multi-element optical systems because it influences subsequent calculations and the overall behavior of the optical system.

Sequential Image Formation

In systems involving more than one refracting surface, the image produced by the first surface typically serves as the object for the following surface. Understanding this sequential process is important because each stage of refraction can alter the characteristics of the image, including its position, orientation, and magnification, ultimately determining the final output of the system.

Transcript

00:01 For this problem on the topic of geometric optics, we are given the refractive index of both ends of a glass rod, and we are told that they are polished to convex hemispherical surfaces.

00:13 We know the radius of curvature of the left end and the radius of curvature at the right end.

00:19 We also know the length of the rod between vertices.

00:23 Now we told that the object for the surface at the left end is an arrow, and we are given the distance to the left of the vertex of the surface where this arrow lies.

00:33 We are also told the height of this arrow, and we are told that it is at right angles to the axis.

00:40 We first want to know what constitutes the object for the surface at the right end of the rod.

00:47 So this will be part a, and we can see here, well, before we do that, we can see that for the first refraction, and we'll label it as such.

01:03 For the first refraction in this problem, we know that the radius of curvature r is equal to positive 6 centimeters, the refractive index, and a is that of a, which is 1, and that for the glass, nb, is 1 .6.

01:29 For the second refraction, we can write down a radius of curvature r, which in this case is minus 12 centimeters, n -a is now that of glass 1 .6, and n -b is simply 1.

01:55 Now, looking at part a of the problem, we want to know what constitutes the object for the surface at the right end of the rod.

02:03 Well, we can see that the image from the left end of the rod actually constitutes as the, or acts as the, object for the right end of the rod and that constitutes our object for the right end of the rod.

02:34 Next in part b we want to know what the object distance is for the surface.

02:42 So we know that n a over s plus n b over s prime is equal to n b minus n a divided by the radius of curvature r.

03:00 So if we substitute our values in, that's 1 over 23 centimeters plus 1 .6 over s prime must equal to 0 .6 over the radius of curvature of 6 centimeters.

03:27 So if we rearrange, we can solve for s prime and we get s prime to be 28 .3.

03:40 Centimeters...

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