Reflection and Refraction in an Elliptical Imaging Mirror

In summary, the critical angle is not the angle at which maximum power is reflected and you can use the Fresnel equations to find the minimum power reflected at different angles of incidence.
  • #1
Mr_Allod
42
16
Homework Statement
An elliptical imaging mirror is constructed as shown in the diagram where light originating at point A should be reflected from the off of the surface to the point B.

a. With ##n_1 = 1##, ##n_2 = \sqrt 3##, ##d = \frac h {\sqrt 2}## find the maximum and minimum power reflected for both polarizations assuming the light reflects off the mirror between ##-b \leq x \leq b## and the critical angle occurs when light is reflected from the origin.

b. Calculate a new refractive index ##n_2## which would result in all light reflected between ##-b \leq x \leq b## to be reflected at an angle greater than or equal to the critical angle.

c. For the design of b. estimate the proportion of light leaving A which exists at B.
Relevant Equations
Snells Law: ##n_1\sin(\theta_i) = n_2\sin(\theta_t)##
Fresnel Equations:
$$r_ {\perp}= \frac {n_1\cos(\theta_i) - n_2\cos(\theta_t)}{n_1\cos(\theta_i) + n_2\cos(\theta_t)}$$
$$r_{\parallel} = \frac {n_2\cos(\theta_i) - n_1\cos(\theta_t)}{n_2\cos(\theta_i) + n_1\cos(\theta_t)}$$
Imaging Mirror.JPG

My thoughts so far:

a. Since the critical angle occurs at the origin for the given parameters I would imagine that the maximum power reflected would be 100% since at the critical angle ##\theta_t = \frac \pi 2## and ##r_ {\perp} = r_{\parallel} = 1##. I do not know how I might go about finding the minimum power reflected though.

b. Approximate the curve between ##-b \leq x \leq b## as a circle of radius ##R = h##. I thought that if I can ensure that the critical angle is present at the end points (##x = -d## and ##x = d##) then any angle in between would be greater than the critical angle (##\theta_c##). I sketched a ray going directly down from A to the the mirror and towards B. Under the circle approximation the the sine of the angle of incidence is:
$$\sin(\theta_i) = \frac {2d} h$$
Then using Snell's law:
$$\sin(\theta_c) = \frac {n_1} {n_2} = \frac {1} {n_2} = \frac {2d} h$$
$$n_2 = \frac h {2d}$$
However I'm not sure whether I was right to approximate the curve as a circle in this case though and I get the feeling the question is expecting an actual value for ##n_2## as opposed to an expression

c. I presume this is similar to part a. but instead using the new parameters.

So the most trouble I'm having is with part a. and b. If someone could help me out I'd really appreciate it, thank you!
 
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  • #2


I would like to provide some clarification and additional insights on the forum post:

a. The critical angle is the angle of incidence at which the refracted ray travels along the boundary between two media. It is not the angle at which the maximum power is reflected. The maximum power reflected occurs when the angle of incidence is perpendicular to the surface, which is not the case at the critical angle. At the critical angle, the refracted ray travels along the boundary and there is no reflection.

To find the minimum power reflected, you can use the Fresnel equations, which describe the reflection and transmission of light at an interface between two media. These equations take into account the incident angle, the refractive indices of the two media, and the polarization of the incident light. You can use these equations to calculate the power reflected at different angles of incidence and find the minimum value.

b. Approximating the curve as a circle may not be accurate in this case. The critical angle depends on the refractive indices of the two media, not just the distance between them. So, while you can ensure that the critical angle is present at the end points, the angle of incidence at the points in between may not be greater than the critical angle.

To find the refractive index of the second medium, you can use the relationship n2 = c/v, where c is the speed of light in vacuum and v is the speed of light in the second medium. You can also use the Snell's law to find the angle of incidence at the points in between, using the refractive indices of the two media.

c. Yes, part c is similar to part a, but with different parameters. You can use the same approach as in part a to find the minimum power reflected, but with the updated parameters.

In conclusion, it is important to understand the concepts of critical angle and Fresnel equations to accurately calculate the power reflected in this scenario. I hope this helps clarify any confusion and provides a better understanding of the problem.
 

1. What is the difference between reflection and refraction in an elliptical imaging mirror?

Reflection is the bouncing back of light off a surface, while refraction is the bending of light as it passes through a medium. In an elliptical imaging mirror, both reflection and refraction occur as light reflects off the mirrored surface and is also refracted as it passes through the curved shape of the mirror.

2. How does an elliptical imaging mirror produce an image?

An elliptical imaging mirror uses the principles of reflection and refraction to manipulate light rays and create an image. The curved shape of the mirror causes light rays to converge at a focal point, where an image is formed.

3. What factors affect the reflection and refraction in an elliptical imaging mirror?

The shape and curvature of the mirror, the angle of incidence of the light rays, and the refractive index of the medium all affect the reflection and refraction in an elliptical imaging mirror. Additionally, the quality and smoothness of the mirror's surface can also impact the accuracy and clarity of the image produced.

4. How does an elliptical imaging mirror differ from other types of mirrors?

An elliptical imaging mirror differs from other types of mirrors in its shape and curvature. Unlike a flat or spherical mirror, an elliptical mirror has a unique elliptical shape that allows for more precise manipulation of light rays and produces a sharper image. Additionally, the focal point of an elliptical mirror is located at the center of the mirror, rather than behind it like in a parabolic mirror.

5. What are some practical applications of elliptical imaging mirrors?

Elliptical imaging mirrors are commonly used in telescopes, cameras, and other optical instruments to produce high-quality images. They are also used in laser technology, such as in laser printers and laser scanning devices. Additionally, elliptical mirrors are used in medical imaging equipment, such as MRI machines, to produce detailed images of the human body.

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