Displacement of virtual image problem

In summary, the conversation is discussing the calculation of the displacement of a virtual object within a glass sphere with a refractive index of 1.5. The goal is to find the image shift, S, as a function of the distance of the object from the center of the sphere, d, when the refracted ray becomes horizontal. The conversation mentions using Snell's law and a diagram to manipulate equations and find expressions for the horizontal distance in terms of angles, d, and S. However, there may be a more efficient method to solve the problem.
  • #1
theguyoo
1
0
Moved from a technical forum, so homework template missing.
An object is within a glass sphere of radius R with a refractive index of 1.5 . I'm trying to calculate the displacement of the virtual object relative to the actual when viewed from the side, such that the refracted ray emanating from the object becomes horizontal. I would like to know S (the image shift) as a function of d, the distance of the object from the center of the sphere.

Diagram: http://imgur.com/dCaTLvq

So far I've tried starting with Snell's law, [itex]sin\alpha = 1.5sin\beta [/itex], and manipulated it to get:
[itex]\frac{\sqrt{R^{2}}-x^{2}}{R}=1.5\frac{x \cdot d}{R\sqrt{R^{2}-2d\sqrt{R^{2}-x^{2}+d^{2}}}}[/itex]where and x is the distance form origin so [itex](S+d)^{2}+x^{2}=R^{2}[/itex] The problem is it's too messy to rearrange to make S the subject, but I'm guessing there's a better way.

Thanks.
 
Physics news on Phys.org
  • #2
Look at the horizontal line from the point where the ray exits the sphere to the apparent position. Can you find two expressions for that distance in terms of the angles, d and S?
 

Related to Displacement of virtual image problem

1. What is the displacement of a virtual image?

The displacement of a virtual image refers to the difference in position between the actual object and its virtual image formed by a mirror or lens.

2. How is the displacement of a virtual image calculated?

The displacement of a virtual image can be calculated using the laws of reflection and refraction, depending on the type of optical element creating the virtual image.

3. Why does the displacement of a virtual image occur?

The displacement of a virtual image occurs because the image is formed by the apparent intersection of light rays, rather than the actual intersection of light rays like in a real image. This can result in a perceived shift in the position of the image.

4. How does the orientation of a mirror affect the displacement of a virtual image?

The orientation of a mirror can affect the displacement of a virtual image by changing the direction of the reflected rays. For example, a concave mirror can produce a virtual image that appears larger and closer to the mirror, while a convex mirror can produce a virtual image that appears smaller and farther away.

5. Can the displacement of a virtual image be controlled or adjusted?

Yes, the displacement of a virtual image can be controlled or adjusted by changing the position or orientation of the optical element, such as a mirror or lens. This can be useful in various applications, such as in telescopes or cameras.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
726
  • Advanced Physics Homework Help
Replies
3
Views
663
  • Introductory Physics Homework Help
Replies
2
Views
911
  • Introductory Physics Homework Help
Replies
8
Views
1K
Replies
11
Views
506
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
7
Views
5K
  • Other Physics Topics
Replies
2
Views
4K
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
2K
Back
Top