Imaging using refraction and its application on prisms

Question

I stumbled upon a problem when trying to explain how a spectrometer with a prism works based on geometrical optics.

I started from the basic example of image formation based on refraction as it is given in the figure below.

What I next did is to see if the distance $h_1$ depends in some way on the angle $i$ and after some calculations I got $$h_1 = h_0 \cdot \sqrt{1-\frac{1-\frac{n_2^2}{n_1^2}}{cos^2(i)}}$$

which implies that it does depend on that angle.

My problem now is related to the idea behind image formation. In order to consider something to be an image, the rays have to intersect into a point, which, based on the above formula, does not happen. Each pair of 2 rays will intersect in a different point.

I tried this online app to check this and the result checked. A caustic appears, not an image in the classical sense.

So the problem is as follows. There is no image formation in the sense that I have mentioned above. However all of us have seen that an image does indeed exist by personal experience. How can this be explained by use of geometrical optics alone?

And on the topic that interested me, that of a prism used as a spectrometer, this implies that the image changes if I change the viewing angle (since this also changes slightly the incident angle). Doesn't this affect the position of spectral lines based on the viewing angle?

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EDIT

As it was requested, I went through the calculation behind the formula in order to check if I got it right. A step-by-step description follows.

$$d = h_0 \cdot tg(i) = h_i \cdot tg(r)$$

from which $h_i$ can be written as

$$h_i = h_0 \frac{sin(i) \cdot cos(r)}{sin(r) \cdot cos(i)}.$$

I replace $sin(r)$ with Snell formula which gives

$$h_i = h_0 \frac{n_2}{n_1} \cdot \frac{cos(r)}{cos(i)}.$$

Rewrite $cos(r)$ using $1=sin^2(r)+cos^2(r)$ and use Snell formula again so

$$h_i = h_0 \frac{n_2}{n_1} \cdot \frac{\sqrt{1-(\frac{n_1}{n_2} sin(i))^2}}{cos(i)}.$$

Get everything under the square root except $h_0$ which gives

$$h_i = h_0 \sqrt{\frac{(\frac{n_2}{n_1})^2-sin^2(i)}{cos^2(i)}}.$$

Change $sin(i)$ into a $cos$ as above and the initial formula is obtained

$$h_i = h_0 \sqrt{\frac{(\frac{n_2}{n_1})^2-1+cos^2(i)}{cos^2(i)}} = h_0 \sqrt{1-\frac{1-(\frac{n_2}{n_1})^2}{cos^2(i)}}.$$

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Answer 1

I think your calculation is correct except for leaving out $h_0$ then putting it back as $h_i$ but the final equation appears correct. I calculated what's multiplying $h_0$ for a few angles in degrees. For 5 and 6 deg the result is 0.6634 and 0.6620 i.e. very close. For 25 deg the result is 0.5688.

I think what you're encountering here is just the aberration at a plane interface. The rays at larger angles do not meet those at smaller angles at a single image location. Refer to Images formed by refraction under the second diagram of the air/water interface where they discuss the small angle approximation. In your equation, if the angle i is small then cos(i) is approximately 1 and what multiplies $h_0$ is just the ratio of the indices. So for small angles, the result is independent of i and the rays appear to come from the same point in the image.

Answer 2

The problem in my setup is that I assumed that I have a point-like source (rays in all directions) and not a plane wave like one. I did a second run on the geometric optics simulator and got this the following image. I used a lens with a point source in its focal point in order to generate parallel rays. I then sent the rays through the prism and used a second lens to focus the beam and it looked quite well focused.

Going back to the initial image in the question, I got to the following "results":

• One has to assume that all the rays are parallel in order to have an image by adding a lens, or, for our day to day experience, the eye lets in rays that are very close to being parallel, while the crystalline lens does the focusing.
• For both cases, images are not formed, where by image I am referring to a point onto which all the rays (or their extensions) intersect. For the point source it is given by the formula, while for the plane wave one needs a lens added to the basic setup.