Concave mirror reflecting a point lamp, that create a wider beam

[Morphus]

Homework Statement

You are building a lamp using a concave mirror and a small lamp (point lamp) which is place on the optical axis at 10 cm from the mirror which have a diameter of 4cm. Knowing that the reflected beam have a 10 degree angle, what is the curvature radius of the mirror.

Relevant Equations

Using the mirror equation

I've look if there was any way to get the "image size" or a ratio to use the Mirror Equation to find the focal length, but nothing. I think it's base on some geometry, but I don't see the relation.

 

[hutchphd]

I believe you wish to have point the source as a "virtual object" and the beam divergence will depend upon the size of the mirror (among other things). You need to do some research I guess....yes this will be geometrical optics.

 

[nasu]

From the angle and diameter of the mirror you can find the position of the (virtual) image.

 

[haruspex]

Homework Statement: You are building a lamp using a concave mirror and a small lamp (point lamp) which is place on the optical axis at 10 cm from the mirror which have a diameter of 4cm. Knowing that the reflected beam have a 10 degree angle, what is the curvature radius of the mirror.
Relevant Equations: Using the mirror equation

I've look if there was any way to get the "image size" or a ratio to use the Mirror Equation to find the focal length, but nothing. I think it's base on some geometry, but I don't see the relation.

It's not a question of image size. The object has no size, so, with an ideal mirror, the image can have no size either.
The subject matter is the dispersion that arises because the mirror is spherical (we have to assume), not parabolic. https://en.wikipedia.org/wiki/Spherical_aberration

 

[nasu]

No, it's not about aberration. The reflected beam is divergent if the source is between the focal point and the mirror. Same will be true for parabolic mirrors.

 

[Morphus]

Thank everyone, yes it's a spherical mirror, I don't think it's about aberration neither but could be wrong. I just don't see the relation of the virtual image position vs angle/diameter.

 

[nasu]

Consider the reflected beam as starting from the virtual image and going through the mirror, back towards the side of the source. The angle between the extreme rays is given.

 

[kuruman]

Knowing that the reflected beam have a 10 degree angle, what is the curvature radius of the mirror.

Is this the angle between the reflected and incident ray or the angle between the reflected and normal to the reflecting surface?

 

[Morphus]

 

[hutchphd]

From the angle and diameter of the mirror you can find the position of the (virtual) image.

You want to find the position of the virtual object which forms from the ray trace (or the lens equation) behind the mirror . That will provide a point source which is "vignetted" (apertured) by the size of the mirror. That is your beam.
The other methods can work but this is a three line solution for a thin lens. Chacun a sa gout

 

[Steve4Physics]

View attachment 324494

I'm not sure about the Post #9 diagram. The lamp is meant produce a cone of light. The virtual image is at the apex of the cone. .If the cone-angle is θ (as shown here: https://undergroundmathematics.org/circles/cones/images/right-cone.svg) then the angle of 10º correspomds to θ or possibly 2θ - the question is unclear.

 

[nasu]

You want to find the position of the virtual object which forms from the ray trace (or the lens equation) behind the mirror . That will provide a point source which is "vignetted" (apertured) by the size of the mirror. That is your beam.
The other methods can work but this is a three line solution for a thin lens. Chacun a sa gout

The object is real, the light source. Behind the mirror is the virtual image.

 

[kuruman]

I'm not sure about the Post #9 diagram. The lamp is meant produce a cone of light. The virtual image is at the apex of the cone. .If the cone-angle is θ (as shown here: https://undergroundmathematics.org/circles/cones/images/right-cone.svg) then the angle of 10º correspomds to θ or possibly 2θ - the question is unclear.

Looking at the drawing, it looks like the 10° is between the optical axis and the outgoing ray which would be the half angle of the cone, $\theta$ in your diagram.

 

[nasu]

View attachment 324494

Is this the image attached to the original question?

 

[Morphus]

Yes... I think there is some line like the 21.8 and 68 degree angle, and the 2 really light gray line. The original question is in french, and it's for my daughter. Some of my Physics and Geometry are quite rusted.

 

[kuruman]

Yes... I think there is some line like the 21.8 and 68 degree angle, and the 2 really light gray line. The original question is in french, and it's for my daughter. Some of my Physics and Geometry are quite rusted.

If you flip the mirror about the vertical, you have an arc (the mirror) that subtends a 20° angle with respect to the center of the of the circle at the virtual image. If the associated chord is 4 units long, how many units is the radius of the circle? Here you will find all the formulas you need with explanations to shake off the rust. Remember that the focal length is half the radius of curvature.

 

[nasu]

The diagram may be something like this. It is not clear if the angle $\theta$ is what is 10 degrees or just half of it. Maybe you post the original text, even if it's in french?

 

[Morphus]

Thank you everyone, will see what I can do with it.

 

[Morphus]

as requested, here is the original text... with the answer

 

[Steve4Physics]

Looking at the drawing, it looks like the 10° is between the optical axis and the outgoing ray which would be the half angle of the cone, $\theta$ in your diagram.

Maybe whoever drew the Post #9 diagram should have marked the angle as '5º' rather than '10º'. The given value of 10º in the question could be the cone's full angle (2θ) or its half-angle (θ). As far as I can see, there's no way to tell which interpretation is correct.

 

[Morphus]

Base on the french text, I assume the 10 degree is the full cone angle since it says "the reflected beam have a 10 degree angle", but I may misinterpreted it

 

[nasu]

With the 10 degree as the full angle I get their answer of 35.5 cm. So, I suppose that this may be the author's intention.
From the right triangle with angle of 5^o (drawing in post 17) you can get the image distance $d_i = \frac{2 cm}{tan 5^o }$ = 22.86 cm which is actually negative (virtual image) according the sign convention which works for the mirror formula I will use in the next paragraph.
From the basic equation $\frac{2}{R}= \frac{1}{d_o}+\frac{1}{d_i}$ with $d_o$=10 cm and $d_i$=-22.9 cm, I get R = 35.5 cm.
This should be actually rounded to 36 cm based on the input data.
When we calculate $d_i$ we neglect the small distance between the vertex of the mirror and the vertical cord. But as the mirror is only 4 cm versus a radius of about 36 cm, the error is of the order of a mm. Anyway, the formula used above assumes paraxial approximation.

 

[Morphus]

Thank