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? (Hint: See Problem $34.69 . )(\mathrm{d})$ 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

Refraction at Spherical Surfaces

This concept involves applying Snell’s law to a curved refractive interface. For spherical surfaces, there is an established formula that relates the object distance, image distance, the refractive indices of the two media, and the radius of curvature. This formulation is central to analyzing how light rays bend when passing from one medium to another at a convex or concave interface.

Sign Conventions in Geometrical Optics

Correctly assigning the signs to distances and radii of curvature is critical in solving imaging problems. The sign conventions determine whether distances are positive or negative based on the orientation of the rays relative to the optical element’s vertex, and they differentiate between concave and convex surfaces. This ensures consistent and accurate application of the imaging equations.

Real and Virtual Objects/Images

In optical systems, an image formed by one element can act as an object for another. Understanding the distinction between real and virtual objects or images is crucial, as it affects the propagation of rays through the system. A real image is formed by actual convergence of rays, while a virtual image is perceived by ray extrapolation, and this concept is essential when analyzing multiple refracting surfaces.

Compound Optical Systems

This concept deals with systems where light undergoes refraction through more than one optical element separated by finite distances. In such systems, the image produced by an initial surface becomes the object for the following surface. Analysis of compound systems requires careful consideration of each component and the separation between them to accurately determine the final image position, orientation, and magnification.

Magnification and Image Height

Magnification quantifies the ratio of the image size to the object size in an optical system and can be determined by relating the distances from the object and image to the refracting surfaces. This concept is important in predicting the final image height and how the orientation of the image (inverted or erect) compares to the original object.

Transcript

00:01 Object image relationship for spherical reflecting surface can be described as n1 upon s plus n2 upon s dash is equal to n2 minus n1 upon r, where n1 and n2 are the refractive, are the refraction index of both of the surfaces.

00:30 S is the object distance from the vertex of the spherical surface, s -dash is the image distance from the vertex of the spherical surface, r is the radius of the spherical surface.

00:45 Now, the sign rules for the variables in the equation are rule number one, sign rule for the object distance.

00:56 When the object is on the same side of the refracting surface as the incoming light object distance as is positive otherwise it is negative rule number two sign rule for the image distance as dash when the image is on the same side of the refracting surface as the outgoing light which is refracted light then the image distance as is positive otherwise it is negative image distance s -dash is positive, otherwise it's negative.

01:36 Rule number three, sign rule for the radius of curvature of a surface which is fatical in our case.

01:47 The rule is when the center of curvature c is on the same side as the outgoing light, which is the reflected light, then the radius of curvature is positive, otherwise it is negative.

02:02 Here, apply the above equation gives us the position of an image formed by refraction of light between a spherical interface of two different mediums.

02:15 To obtain this position, we need to get the radius of the spherical interface and the two refraction indexes of the mediums and the object position.

02:30 Here note that we can use this formula for a plain interface when we put r equals to infinity as if the line of interface is considered a part of sphere which has a radius of infinity.

02:49 Now let's talk about the party.

02:53 Here the image caused by the refraction on the left end of the role is the object.

03:01 For the right end of the road.

03:07 Now, part b, we have refractive index of air n1 is 1 and for the roar, refractive index is 1 .6 which is n2.

03:27 Our left is 6 cm is positive as the surface is convex and the radius in direction of outgoing rays.

03:37 So let's put the values in this equation and we can write this equation as well in the form of s dash.

03:48 So s dash is equal to n2 upon n2 minus n1 divided by r minus n1 upon s is equal to 1 .6 upon 1 .6 minus 1 divided by 6 minus 1 over 23.

04:25 So we will get s dash is equal to 28 .3 centimeter to the right of the left end and to right and 40 minus 28 .3 centimeter.

04:57 So we will get 11 .7 centimeter to the left of the right end.

05:12 Now in part c, the object is in the direction of the outgoing rays from the left side.

05:21 So it is a real object.

05:36 First, let's mention this is our part b.

05:42 This is part c and the object is real object.

05:46 Additionally, s -dash is positive in part b, which confirms that.

05:54 Now in part b, the second refraction by the right end, the object is 11 .7 cm from the left, as shown in the parting.

06:09 So here, s is equal to 11 .7 centimeter and r is equal to minus 12 centimeter is negative as the surface is convex, and the radius is opposite direction of outgoing rays.

06:28 So again we can calculate s -tash in this case with the help of same formula that we have used double.

06:37 So s -dash is equal to n2 which is 1 divided by n2 minus n1.

06:44 So 1 minus 1 .6 upon r which is minus 12 minus n1 is 1 .6.

06:57 Upon s is 11 .7.

07:03 So we will get s -dash is equal to minus 11 .52 centimeter...

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