How Do Refraction and Radii Affect Image Formation in a Glass Rod?

[bodensee9]

Homework Statement

Hello: I am wondering if someone can help with the following?
A rod that is 96 cm long is made of glass with index of refraction of 1.6. The rod has its end ground to convex spherical surfaces with radii of 8 cm and 16 cm. An object is in air on the long axis of the rod 20 cm from the end with the 8 cm radius. Find the image distance due to refraction at the 8 cm surface. Find also the position of the final image due to refraction at both surface.

The Attempt at a Solution

I know the first one. I will have: 1/20 + 1.6/s = (1.6-1)/8. So then I get 64 cm. This means that the image of the object is 64 cm inside the glass tube. I am not sure about the second one. I know that the image from the 8 cm mirror will serve as the object for the 16 cm mirror. So then do I have that the image from the 8 cm mirror is (96-64) from the end of the 16 cm mirror? I am not sure about the signs. I know that distance is positive if it is on the light incident side. So would distance from the image of 8 cm to the radius of the 16 cm mirror be negative because the light incident side would come from the outside of the glass tube? And wouldn't the indices of refraction be the same? Since the equation is n1/s1+n2/s2 = (n2-n1)/r. Would I have 1.6/32 + 1/s2 = (-.6)/16? I think the image of the 16cm will be on the outside of the glass tube. Thanks

 

[anathema]

for any help!

Thank you for your post. I can help you with the second part of your problem.

You are correct in thinking that the image from the 8 cm mirror will serve as the object for the 16 cm mirror. However, the distance from the image of 8 cm to the radius of the 16 cm mirror should be positive, since it is on the light incident side. This is because light travels from the object to the mirror, so the distance should be measured in the direction of the light's travel.

Also, the indices of refraction will not be the same for the second part. This is because the light is now traveling from the glass (with an index of refraction of 1.6) to the air (with an index of refraction of 1). So the equation you should use is n1/s1 + n2/s2 = (n2-n1)/r, where n1 = 1.6, n2 = 1, s1 = 64 cm, and r = 16 cm.

Hope this helps. Let me know if you have any further questions or if you need clarification. Good luck with your problem!

 

[kerimek]

for any help!I would first commend you on your attempt at solving the problem and using the correct equations. However, there are a few things that need to be clarified in order to fully solve the problem.

Firstly, in the equation 1/20 + 1.6/s = (1.6-1)/8, the 1/20 term represents the distance from the object to the first surface, while the 1.6/s term represents the distance from the first surface to the image. Therefore, the 64 cm value that you obtained represents the distance from the object to the image, not the image distance itself. The image distance can be found by rearranging the equation to solve for s, which would give you s = 32 cm.

Secondly, when considering the second surface, you are correct in thinking that the image from the first surface will serve as the object for the second surface. However, the distance from the image to the second surface should be calculated as (96-32) cm, not (96-64) cm. This is because the image from the first surface is located 32 cm from the first surface, not 64 cm.

In terms of signs, you are correct in thinking that distances on the incident side of the light are positive and distances on the opposite side are negative. However, in this problem we are dealing with refraction, which means that the light rays are bending as they pass through the glass. Therefore, the image distance from the second surface should be considered negative, since it is on the opposite side of the light rays. This would result in the equation 1.6/-64 + 1/s2 = (-.6)/16.

Lastly, the indices of refraction for the two surfaces should be different, since the first surface has a radius of 8 cm while the second surface has a radius of 16 cm. Therefore, the equation should be 1.6/32 + 1/s2 = (-.6)/16. This would give you a final image distance of s2 = -32 cm, which means that the image is located 32 cm on the opposite side of the second surface.

I hope this helps clarify some of the confusion and allows you to fully solve the problem. Keep up the good work!