comparing angle in measurement for concave mirror
$begingroup$
From the above picture we can prove easily $ CF = XF$ because CFX is a isosceles triangle. Now if we move the point X more near near P, it will give the same results(I mean we will get again $ CF = XF$) but by drawing the figure on paper, I have measured that XF= CF are not equal always. What is wrong here?
geometry
edited Dec 23 '18 at 21:02
Glorfindel
3,41981830
asked May 24 '13 at 20:32
Math_guyMath_guy
133
$endgroup$
migrated from physics.stackexchange.com May 24 '13 at 21:07
This question came from our site for active researchers, academics and students of physics.
|
show 7 more comments
$begingroup$
From the above picture we can prove easily $ CF = XF$ because CFX is a isosceles triangle. Now if we move the point X more near near P, it will give the same results(I mean we will get again $ CF = XF$) but by drawing the figure on paper, I have measured that XF= CF are not equal always. What is wrong here?
geometry
edited Dec 23 '18 at 21:02
Glorfindel
3,41981830
asked May 24 '13 at 20:32
Math_guyMath_guy
133
$endgroup$
migrated from physics.stackexchange.com May 24 '13 at 21:07
This question came from our site for active researchers, academics and students of physics.
$begingroup$
a-levelphysicstutor.com/images/optics/ccmrrs-r-2f.jpg
$endgroup$
– Physics_guy
May 24 '13 at 20:33
5
$begingroup$
What is the nature of the particular curve? I would have guessed that it’s supposed to be a parabola. What gives you the idea that the curve is accurately drawn?
$endgroup$
– Lubin
May 24 '13 at 21:19
1
$begingroup$
if you move the point X more near near P, it will give the same results iff CF=FP..
$endgroup$
– mle
May 24 '13 at 21:28
$begingroup$
why was this migrated to MSE? This is obviously a question on geometrical optics, it belongs here (geometry) as much as it belongs there (optics).
$endgroup$
– user31280
May 24 '13 at 21:31
1
$begingroup$
Here you go.
$endgroup$
– newbie
May 25 '13 at 0:39
|
show 7 more comments
$begingroup$
From the above picture we can prove easily $ CF = XF$ because CFX is a isosceles triangle. Now if we move the point X more near near P, it will give the same results(I mean we will get again $ CF = XF$) but by drawing the figure on paper, I have measured that XF= CF are not equal always. What is wrong here?
geometry
edited Dec 23 '18 at 21:02
Glorfindel
3,41981830
asked May 24 '13 at 20:32
Math_guyMath_guy
133
$endgroup$
From the above picture we can prove easily $ CF = XF$ because CFX is a isosceles triangle. Now if we move the point X more near near P, it will give the same results(I mean we will get again $ CF = XF$) but by drawing the figure on paper, I have measured that XF= CF are not equal always. What is wrong here?
geometry
geometry
edited Dec 23 '18 at 21:02
Glorfindel
3,41981830
asked May 24 '13 at 20:32
Math_guyMath_guy
133
edited Dec 23 '18 at 21:02
Glorfindel
3,41981830
asked May 24 '13 at 20:32
Math_guyMath_guy
133
edited Dec 23 '18 at 21:02
Glorfindel
3,41981830
edited Dec 23 '18 at 21:02
Glorfindel
3,41981830
edited Dec 23 '18 at 21:02
Glorfindel
3,41981830
3,41981830
asked May 24 '13 at 20:32
Math_guyMath_guy
133
asked May 24 '13 at 20:32
Math_guyMath_guy
133
asked May 24 '13 at 20:32
Math_guyMath_guy
133
133
migrated from physics.stackexchange.com May 24 '13 at 21:07
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com May 24 '13 at 21:07
This question came from our site for active researchers, academics and students of physics.
$begingroup$
a-levelphysicstutor.com/images/optics/ccmrrs-r-2f.jpg
$endgroup$
– Physics_guy
May 24 '13 at 20:33
5
$begingroup$
What is the nature of the particular curve? I would have guessed that it’s supposed to be a parabola. What gives you the idea that the curve is accurately drawn?
$endgroup$
– Lubin
May 24 '13 at 21:19
1
$begingroup$
if you move the point X more near near P, it will give the same results iff CF=FP..
$endgroup$
– mle
May 24 '13 at 21:28
$begingroup$
why was this migrated to MSE? This is obviously a question on geometrical optics, it belongs here (geometry) as much as it belongs there (optics).
$endgroup$
– user31280
May 24 '13 at 21:31
1
$begingroup$
Here you go.
$endgroup$
– newbie
May 25 '13 at 0:39
|
show 7 more comments
$begingroup$
a-levelphysicstutor.com/images/optics/ccmrrs-r-2f.jpg
$endgroup$
– Physics_guy
May 24 '13 at 20:33
5
$begingroup$
What is the nature of the particular curve? I would have guessed that it’s supposed to be a parabola. What gives you the idea that the curve is accurately drawn?
$endgroup$
– Lubin
May 24 '13 at 21:19
1
$begingroup$
if you move the point X more near near P, it will give the same results iff CF=FP..
$endgroup$
– mle
May 24 '13 at 21:28
$begingroup$
why was this migrated to MSE? This is obviously a question on geometrical optics, it belongs here (geometry) as much as it belongs there (optics).
$endgroup$
– user31280
May 24 '13 at 21:31
1
$begingroup$
Here you go.
$endgroup$
– newbie
May 25 '13 at 0:39
$begingroup$
a-levelphysicstutor.com/images/optics/ccmrrs-r-2f.jpg
$endgroup$
– Physics_guy
May 24 '13 at 20:33
$begingroup$
a-levelphysicstutor.com/images/optics/ccmrrs-r-2f.jpg
$endgroup$
– Physics_guy
May 24 '13 at 20:33
5
5
$begingroup$
What is the nature of the particular curve? I would have guessed that it’s supposed to be a parabola. What gives you the idea that the curve is accurately drawn?
$endgroup$
– Lubin
May 24 '13 at 21:19
$begingroup$
What is the nature of the particular curve? I would have guessed that it’s supposed to be a parabola. What gives you the idea that the curve is accurately drawn?
$endgroup$
– Lubin
May 24 '13 at 21:19
1
1
$begingroup$
if you move the point X more near near P, it will give the same results iff CF=FP..
$endgroup$
– mle
May 24 '13 at 21:28
$begingroup$
if you move the point X more near near P, it will give the same results iff CF=FP..
$endgroup$
– mle
May 24 '13 at 21:28
$begingroup$
why was this migrated to MSE? This is obviously a question on geometrical optics, it belongs here (geometry) as much as it belongs there (optics).
$endgroup$
– user31280
May 24 '13 at 21:31
$begingroup$
why was this migrated to MSE? This is obviously a question on geometrical optics, it belongs here (geometry) as much as it belongs there (optics).
$endgroup$
– user31280
May 24 '13 at 21:31
1
1
$begingroup$
Here you go.
$endgroup$
– newbie
May 25 '13 at 0:39
$begingroup$
Here you go.
$endgroup$
– newbie
May 25 '13 at 0:39
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The geometry in the diagram is true no matter what the figure of the mirror is; that is, $triangle XFC$ is isosceles. What is not true is that both $C$ and $F$ are independent of $X$. In what follows, $kappa$ is the curvature of the curve at $P$.
If we wish to fix $F$ independent of $X$, then the curve needs to be a parabola. However, in that case, $C$ is dependent on $X$.
$hspace{2cm}$
If we wish to fix $C$ independent of $X$, then the curve needs to be a circle. However, in that case, $F$ is dependent on $X$.
$hspace{2cm}$
Note that as $rto0$, $PCto2PF$.
answered May 25 '13 at 14:53
robjohn♦robjohn
268k27308633
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The geometry in the diagram is true no matter what the figure of the mirror is; that is, $triangle XFC$ is isosceles. What is not true is that both $C$ and $F$ are independent of $X$. In what follows, $kappa$ is the curvature of the curve at $P$.
If we wish to fix $F$ independent of $X$, then the curve needs to be a parabola. However, in that case, $C$ is dependent on $X$.
$hspace{2cm}$
If we wish to fix $C$ independent of $X$, then the curve needs to be a circle. However, in that case, $F$ is dependent on $X$.
$hspace{2cm}$
Note that as $rto0$, $PCto2PF$.
answered May 25 '13 at 14:53
robjohn♦robjohn
268k27308633
$endgroup$
add a comment |
$begingroup$
The geometry in the diagram is true no matter what the figure of the mirror is; that is, $triangle XFC$ is isosceles. What is not true is that both $C$ and $F$ are independent of $X$. In what follows, $kappa$ is the curvature of the curve at $P$.
If we wish to fix $F$ independent of $X$, then the curve needs to be a parabola. However, in that case, $C$ is dependent on $X$.
$hspace{2cm}$
If we wish to fix $C$ independent of $X$, then the curve needs to be a circle. However, in that case, $F$ is dependent on $X$.
$hspace{2cm}$
Note that as $rto0$, $PCto2PF$.
answered May 25 '13 at 14:53
robjohn♦robjohn
268k27308633
$endgroup$
add a comment |
$begingroup$
The geometry in the diagram is true no matter what the figure of the mirror is; that is, $triangle XFC$ is isosceles. What is not true is that both $C$ and $F$ are independent of $X$. In what follows, $kappa$ is the curvature of the curve at $P$.
If we wish to fix $F$ independent of $X$, then the curve needs to be a parabola. However, in that case, $C$ is dependent on $X$.
$hspace{2cm}$
If we wish to fix $C$ independent of $X$, then the curve needs to be a circle. However, in that case, $F$ is dependent on $X$.
$hspace{2cm}$
Note that as $rto0$, $PCto2PF$.
answered May 25 '13 at 14:53
robjohn♦robjohn
268k27308633
$endgroup$
The geometry in the diagram is true no matter what the figure of the mirror is; that is, $triangle XFC$ is isosceles. What is not true is that both $C$ and $F$ are independent of $X$. In what follows, $kappa$ is the curvature of the curve at $P$.
If we wish to fix $F$ independent of $X$, then the curve needs to be a parabola. However, in that case, $C$ is dependent on $X$.
$hspace{2cm}$
If we wish to fix $C$ independent of $X$, then the curve needs to be a circle. However, in that case, $F$ is dependent on $X$.
$hspace{2cm}$
Note that as $rto0$, $PCto2PF$.
answered May 25 '13 at 14:53
robjohn♦robjohn
268k27308633
edited May 26 '13 at 0:17
answered May 25 '13 at 14:53
robjohn♦robjohn
268k27308633
answered May 25 '13 at 14:53
robjohn♦robjohn
268k27308633
answered May 25 '13 at 14:53
robjohn♦robjohn
268k27308633
268k27308633
add a comment |
add a comment |
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$begingroup$
a-levelphysicstutor.com/images/optics/ccmrrs-r-2f.jpg
$endgroup$
– Physics_guy
May 24 '13 at 20:33
5
$begingroup$
What is the nature of the particular curve? I would have guessed that it’s supposed to be a parabola. What gives you the idea that the curve is accurately drawn?
$endgroup$
– Lubin
May 24 '13 at 21:19
1
$begingroup$
if you move the point X more near near P, it will give the same results iff CF=FP..
$endgroup$
– mle
May 24 '13 at 21:28
$begingroup$
why was this migrated to MSE? This is obviously a question on geometrical optics, it belongs here (geometry) as much as it belongs there (optics).
$endgroup$
– user31280
May 24 '13 at 21:31
1
$begingroup$
Here you go.
$endgroup$
– newbie
May 25 '13 at 0:39