# Real image drawn in case of concave mirror is bit off

I am trying to draw real image of object in front of concave mirror. Case is when object is lying on the Center of curvature.

What I have done so far

There are two rays

1. A ray incident on the mirror parallel to the principal axis reflects through the focal point.
2. Another ray passing through the focal point reflects back parallel to the principal axis.

Actual Result

if you kindly notice the image below, the image of real object is lying little bit right side from Center of Curvature.

Expected Result

The real image should be present just below the Center of curvature and of same size of object. but there is bit mismatching as shown in the image.

Statistics

My calculations are based on following.

From Pole to Focal point = 4cm
From Focal point to Center of Curvature = 4cm
Curve drawn was 8cm away from Center of Curvature.

Am I missing anything?

You ray is not paraxial, so you are experiencing spherical aberration. See this image from the article

• OP is not ray tracing with calculations but drawing rays based on paraxial optics so the image should be below C. See my answer. Commented Apr 14 at 13:06
• I just studied paraxial and found that an incident ray that is close to the pole is called paraxial ray. If this understanding is correct then there is no paraxial ray in my case. Commented Apr 14 at 13:59
• Paraxial means that you can use small-angle approximations, so the rays should be almost parallel to the optical axis. Your incident rays are parallel, but the reflected ones are not. It is not clear to me how you are constructing your figure though so my answer may not be relevent --- as @Not_Einstein has said. I was assuming that you were making the angle of incidence on the mirror equal the angle of reflection --- rather than using principal rays. Commented Apr 14 at 14:15

It appears that your diagram is not drawn accurately. The distance CF is less than the distance FP. Try it again with a more accurate diagram.

• CF and FP both are 4cm and it might appear not equal maybe because while taking the photo from Android phone, it does show some part is emerging out from right side. Commented Apr 14 at 13:53

You can not rely on graph, particularly hand-drawn graph, for facts, but only intuitions, which shoud be proven or disproven theoritically.

Especially in your case, because of the approximation your model makes. Only parabolic mirrors have focal points. We aproximate one as parabolic if it is only a small part around the axis.

In your case, you missed the condition that it is a spherical mirror, hence the normals at the reflection points go through C, the center of curvature. And the normals are both radius of the sphere.

(Previous answer edited after realisation of a mistake. Comments below were for the previous version.)

I agree with other answers about the invalidity of paraxial approximations in this case. However, there are two things to note:

1. If you draw the normal at the point of incidence and ensure that the angles of incidence and reflection are equal, all rays parallel to the principal axis will not pass through the focus. This results in spherical aberration.
2. Even if you forcefully draw the lines passing through the focus after reflection, assuming paraxial approximation, your image location will be different from what you expect, since the point of reflection from the mirror is still the same, and this affects the angle at which the ray passes through the focus.

Paraxial approximation is equivalent to assuming that the distance from the object to the mirror is a constant irrespective of the height. For finite size objects, this means that the dimensions of the object is much smaller than the radius of curvature of the mirror. (I don't like the description of "rays close to the principal axis" - How close?)

If you make your mirror almost flat, instead of drawing a proper arc with $$C$$ as centre, you will get what you expect, with the image at the same distance as the object. However, it will not be drawn to scale, but you could say that the scale is different in the $$x$$- and $$y$$-dimensions.

• That is not the only possible explanation. Another reason could be that the rays are not paraxial, leading to spherical aberration. Commented Apr 18 at 12:51
• @Stuti Nope, that's wrong. The rays here are drawn exactly according to the paraxial approximation, which is why the rays parallel to the principal axis pass through the focus. To demonstrate the effects of spherical aberration, you have to explicitly offset the reflected rays from the focus depending on the distance of the incident parallel ray from the principal axis, ensuring that the angles of incidence and reflection are equal. If the current diagram is drawn according to scale, one can use similarity of triangles and argue that the image should be at the same distance as the object. Commented Apr 18 at 14:05
• @Stuti If you draw a proper normal at the point of incidence and make sure that the angles of incidence and reflection are equal, then it would not pass through the focus, which is drawn midway between the centre of curvature and the pole. Commented Apr 18 at 14:09
• The rays here are drawn exactly according to the paraxial approximation - that is exactly what I'm talking about. How does OP know that the angle subtended by the object on the pole is small enough that they can assume the rays to be paraxial? If you draw a proper normal at the point of incidence and make sure that the angles of incidence and reflection are equal, then it would not pass through the focus - that is why they are NOT paraxial. Commented Apr 18 at 14:16
• @Stuti I'm sorry, I realised the mistake I made. I edited the answer. Commented Apr 18 at 14:50
1. A ray incident on the mirror parallel to the principal axis reflects through the focal point.
2. Another ray passing through the focal point reflects back parallel to the principal axis.

Neither of these two statements are accurate and these 'rules' are more formally known as small angle approximations. Statement (1) contains the phrase "the focal point" which is poorly defined for a circular cross section concave mirror. This is clearly demonstrated in the two diagrams below. The parallel ray coming from the tall object has its focus in a different location to that of a parallel ray coming from a shorter object, so there is no single focal point for the parallel rays. However, it can be seen that when the parallel rays are closer to the main axis and the angles are smaller, the closer the focal point gets to R/2 , the closer the image location gets to R, which is why they are called small angle approximations.

It can also be seen that statement (2) is not accurate because it is it clear that the ray coming from the tall object that passes through the 'focus', does not return parallel to the main axis. It is less obvious for the shorter object, but it is still not quite parallel and only really approaches being parallel when the the angles approach zero. The diagrams also confirm your conclusion that the formed image is to the right of the centre of curvature. The above diagrams are drawn by ignoring the approximate rules and using the main rule that can always be relied on for reflections in curved mirrors, which is that the angle of reflection is always equal to the angle of incidence relative to the normal (the dashed lines in the diagrams) of the mirror, which is a line perpendicular to the tangent of the curve at the reflection point

The diagram on left above shows that if the object is moved to a point where it is at the same horizontal location as the formed image, then we have what looks like a more normal focal point with the ray passing through the focal point returning parallel to the main axis. The ray reflected from the point of the mirror on the main axis now also passes through the image point (which you may have noticed did not happen in the first image at the very top). However, the image on the right (above) demonstrates there are still reflection points on the mirror, where the rays are not directed exactly at the image point, so there is never a completely blur free image formed of any point on the object. This is a limitation of using curved mirrors to form images.

The final two diagrams analyse the behaviour of a parabolic mirror for comparison purposes. The parabolic mirror behaves a bit more like you would expect. The focal point remains in the same place when the height of the incoming parallel rays are altered. Rays passing through the focal point do indeed always return parallel to the main axis. However, the parabolic mirror is not perfect either. It can be seen in the diagram on the left, when the parallel rays from the object are far from the main axis, that the green ray does not end up at the image point. It can also be seen that the location of the formed image is not exactly in the same place for high and low parallel rays, but this effect is much less pronounced that for the the circular mirror. These last two effects mean a parabolic mirror does not form a perfect image either of near objects. (For far away objects like planets and stars, the angles are very small and the errors tend towards negligible.)

In summary, your ray tracing diagram is not quite accurate, because you have assumed the ray that passes through the focal point returns parallel to the main axis, which I have shown is not always the case by using the principle that reflected rays always have the same angle as the incident ray relative to the normal of the mirror surface. However, the error is small and by either analysis, the image is indeed to the right of the centre of curvature.

You can verify all this for yourself by using this GeoGebra applet. Any of the green points can be dragged to change the parameters.

As the formula derived on paper uses paraxial approximations and also curve-pole approximation, How about we just go to basics,Use laws of reflection by drawing normal for each ray you throw from the object, That will Give you how real Rays behave.

If you just want to stick to getting image at centre of curvature, Try increasing the radius of Curve and decreasing the height of object, You will approximately get Image at centre of Curvature, Because these were the ways the formula was derived using approximations.