# Is the Solution for Rotation of Spherical Mirrors Incorrect?

• Aurelius120
Aurelius120
Homework Statement
Find the position of final image
Relevant Equations
Lens formula: $$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$
Mirror Formula: $$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$
Mirror Magnification = ##\frac{-v}{u}##
Lens Magnification =##\frac{v}{u}##

I think the given solution is wrong.
The lens forms image at ##(+75,0)## which is ##25 cm## from pole of the convex mirror which acts as virtual object for mirror.

It is true that the reflected ray is rotated by ##2\theta## as in case of plane mirror. Rotation of Spherical Mirrors
But that doesn't necessarily mean that the image is rotated by ##2\theta##

Since the mirror is tilted the principal axis is tilted. It is a spherical mirror so the point object should be treated as the tip of an extended object and the height and position of head of extended object should give coordinates of point image.

According to this diagram:
$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\implies \frac{1}{v}+\frac{2}{25\sqrt{3}}=\frac{1}{50}$$
Therefore $$v=\frac{50\sqrt{3}}{\sqrt{3}-4}$$ ## \text{ v is along tilted axis and measured from pole of mirror}##

Then from the formula of magnification:
$$\frac{I}{25/2}=\frac{-v}{25\sqrt{3}/2}\implies I=\frac{50}{4-\sqrt{3}}$$

Solving this triangle,
$$(|v|-I\tan 30)×\cos 30=50-x$$
$$y=\frac{I}{\cos 30}+(|v|-I\tan 30)×\sin 30$$
$$x=\frac{50(3-\sqrt{3}}{4-\sqrt{3}}\text{ ; } y=\frac{50\sqrt{3}}{4-\sqrt{3}}$$
These equations give the x and y coordinates of the image.

The answer does not match. There is yet another mistake in the solution, I think. I believe that this image is not the final image. The final image is formed after refraction of this image by lens.

So am I horribly misunderstanding the question? Am I terribly mistaken or is the solution wrong this time??
How to solve it correctly?

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Here is a similar Question that uses this method in tilted mirror

@haruspex If you see this, What do you think?

Aurelius120 said:
@haruspex If you see this, What do you think?
I've never been confident in dealing with mirror/lens problems.
What worries me here is that the tilted mirror creates a setup akin to an untilted mirror but with the object being similar in size the mirror radius. The usual approximations cannot be used.
I'll see if I can find a method that convinces me.

haruspex said:
I've never been confident in dealing with mirror/lens problems.
What worries me here is that the tilted mirror creates a setup akin to an untilted mirror but with the object being similar in size the mirror radius. The usual approximations cannot be used.
I'll see if I can find a method that convinces me.
So who do you think is an optics mentor here that I can tag?

Lnewqban said:
Consider that the rays hitting the convex mirror are not parallel, as the imaging is past the 30 cm focal point of the convex lens.
How does that change things? The image formed by lens at ##(+75,0)## acts as a virtual point object at ##(25,0)## for convex mirror. Which I am treating as the head of an extended objecct because of tilted axis of the mirror

Lnewqban said:
This may be able to help you:
https://www.texasgateway.org/resource/86-image-formation-mirrors

Consider that the rays hitting the convex mirror are not parallel, as the imaging is past the 30 cm focal point of the convex lens.
After reading that article I came up with these diagrams:

For small aperture, tangents approach the arcs
Therefore:

I haven't solved these triangles but I think the results will be similar from the looks of it. Please help solving this

Does the answer match my solution or the official one?

I don’t understand your first diagram in post #8.
Where is the image from the lens? Where does the ray reaching I from the line CO come from?

haruspex said:
I don’t understand your first diagram in post #8.
Where is the image from the lens? Where does the ray reaching I from the line CO come from?
O is the image from the lens (virtual object for mirror)
I is the image of O
Since C is center of curvature, ray of light along CO is case of normal incidence and retraces its path after reflection.

Aurelius120 said:
O is the image from the lens (virtual object for mirror)
I is the image of O
Since C is center of curvature, ray of light along CO is case of normal incidence and retraces its path after reflection.
Ok, that works. I was confused because I expected to see two points (or a line segment) representing the image from the lens. The point you found only shows where the image ends up for that part of the object which lies on the axis of the lens.

haruspex said:
Ok, that works. I was confused because I expected to see two points (or a line segment) representing the image from the lens. The point you found only shows where the image ends up for that part of the object which lies on the axis of the lens.
But it is a small object or point object? Without height of object given in question, how can I solve otherwise?

In my first post I treated that point object as the tip of an extended object, therefore the line segment

Aurelius120 said:
But it is a small object or point object?
It is described as a small object, and illustrated as a short arrow. You don’t need to know an actual height to figure out the orientation of the image and its magnification, but I see you are not asked for those, so forget my post #11.

The only problem I have with your solution in post #8 is spherical aberration. Usually, one only considers rays close to the axis. Your two rays are a wide angle apart.
As the picture here shows…
https://en.wikipedia.org/wiki/Spherical_aberration
the sharpest image lies on the line through the object and the centre of curvature. That implies the right combination of rays to consider is your IOC ray and another nearby, not the one along the axis of the lens.

Update:
Doh! That is the answer. The final image is the same as if the object were at O and the sphere were mirrored on the inside. So treat the mirror axis as COI and apply the standard formula. Do not use the ray along the lens axis.

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haruspex said:
Doh! That is the answer. The final image is the same as if the object were at O and the sphere were mirrored on the inside. So treat the mirror axis as COI and apply the standard formula. Do not use the ray along the lens axis.
The only problem is : I can't seem to find the angle between IOC and x-axis(lens axis). So I can't find object distance. I think it's approximately ##30°##

Aurelius120 said:
The only problem is : I can't seem to find the angle between IOC and x-axis(lens axis). So I can't find object distance. I think it's approximately ##30°##
Call the point where the mirror intersects the lens' axis P.
You know OP, CP and angle CPO. That is enough to find all sides and angles of triangle OPC.

@haruspex

From this figure
$$PR=100\cos 30=50\sqrt{3}; CR=50$$
$$OC=\sqrt{50^2+(50\sqrt{3}-25)^2}$$
$$u=100-\sqrt{50^2+(50\sqrt{3}-25)^2}=100-25\sqrt{4+(2\sqrt 3-1)^2}$$
That seems a little too complex

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Aurelius120 said:
@haruspex
View attachment 346854
From this figure
$$PR=100\cos 30=50\sqrt{3}; CR=50$$
$$OC=\sqrt{50^2+(50\sqrt{3}-25)^2}$$
$$u=100-\sqrt{50^2+(50\sqrt{3}-25)^2}=100-25\sqrt{4+(2\sqrt 3-1)^2}$$
That seems a little too complex
It is right, but there is a shorter route. Using the cosine rule, ##OC^2=PC^2+PO^2-2.PC.OC.\cos(OPC)##.

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