Mixing
geometry - find the path a light ray must take to reach a destination with one bounce off a mirror
$begingroup$
this seems like it should be simple, but i've run out of leads with a similar-triangles approach, and the algebraic approaches seem pretty daunting, so i'm asking for help.
i'm working on a simulation where a ray of light reaches a point by bouncing once off a planar mirror. i know the position of the source, the destination point, and a point & normal of the mirror. (or any other convenient means of describing the mirror) For simplicity i'm happy to assume the mirror is infinite in extent.
as i mentioned, a geometric solution via similar-triangles and so on has not revealed itself to me, but i'm good at missing those.
for the algebraic approach i've set up a system of equations constraining the on-mirror point to 1) lie on the line of the mirror and 2) the angle between the incoming ray and normal and the angle between the outgoing ray and the normal are equal, by asserting that those Dot-products are equal. I haven't carried that math through to its end because it looks pretty monstrous because the thing i'm trying to solve for appears in so many places including inside many radicals.
I'm not saying i can't solve this, but it looks like a lot of work and if there's a better approach i'd love some advice.
Here's my sketch of the situation. Apologies if i mis-use terminology!
Speaking of terminology, is there a word for this point ? The point of reflection ? The point of incidence ?
geometry reflection
asked Jan 29 at 17:09
orion elenzilorion elenzil
1153
$endgroup$
add a comment |
$begingroup$
this seems like it should be simple, but i've run out of leads with a similar-triangles approach, and the algebraic approaches seem pretty daunting, so i'm asking for help.
i'm working on a simulation where a ray of light reaches a point by bouncing once off a planar mirror. i know the position of the source, the destination point, and a point & normal of the mirror. (or any other convenient means of describing the mirror) For simplicity i'm happy to assume the mirror is infinite in extent.
as i mentioned, a geometric solution via similar-triangles and so on has not revealed itself to me, but i'm good at missing those.
for the algebraic approach i've set up a system of equations constraining the on-mirror point to 1) lie on the line of the mirror and 2) the angle between the incoming ray and normal and the angle between the outgoing ray and the normal are equal, by asserting that those Dot-products are equal. I haven't carried that math through to its end because it looks pretty monstrous because the thing i'm trying to solve for appears in so many places including inside many radicals.
I'm not saying i can't solve this, but it looks like a lot of work and if there's a better approach i'd love some advice.
Here's my sketch of the situation. Apologies if i mis-use terminology!
Speaking of terminology, is there a word for this point ? The point of reflection ? The point of incidence ?
geometry reflection
asked Jan 29 at 17:09
orion elenzilorion elenzil
1153
$endgroup$
add a comment |
$begingroup$
this seems like it should be simple, but i've run out of leads with a similar-triangles approach, and the algebraic approaches seem pretty daunting, so i'm asking for help.
i'm working on a simulation where a ray of light reaches a point by bouncing once off a planar mirror. i know the position of the source, the destination point, and a point & normal of the mirror. (or any other convenient means of describing the mirror) For simplicity i'm happy to assume the mirror is infinite in extent.
as i mentioned, a geometric solution via similar-triangles and so on has not revealed itself to me, but i'm good at missing those.
for the algebraic approach i've set up a system of equations constraining the on-mirror point to 1) lie on the line of the mirror and 2) the angle between the incoming ray and normal and the angle between the outgoing ray and the normal are equal, by asserting that those Dot-products are equal. I haven't carried that math through to its end because it looks pretty monstrous because the thing i'm trying to solve for appears in so many places including inside many radicals.
I'm not saying i can't solve this, but it looks like a lot of work and if there's a better approach i'd love some advice.
Here's my sketch of the situation. Apologies if i mis-use terminology!
Speaking of terminology, is there a word for this point ? The point of reflection ? The point of incidence ?
geometry reflection
asked Jan 29 at 17:09
orion elenzilorion elenzil
1153
$endgroup$
this seems like it should be simple, but i've run out of leads with a similar-triangles approach, and the algebraic approaches seem pretty daunting, so i'm asking for help.
i'm working on a simulation where a ray of light reaches a point by bouncing once off a planar mirror. i know the position of the source, the destination point, and a point & normal of the mirror. (or any other convenient means of describing the mirror) For simplicity i'm happy to assume the mirror is infinite in extent.
as i mentioned, a geometric solution via similar-triangles and so on has not revealed itself to me, but i'm good at missing those.
for the algebraic approach i've set up a system of equations constraining the on-mirror point to 1) lie on the line of the mirror and 2) the angle between the incoming ray and normal and the angle between the outgoing ray and the normal are equal, by asserting that those Dot-products are equal. I haven't carried that math through to its end because it looks pretty monstrous because the thing i'm trying to solve for appears in so many places including inside many radicals.
I'm not saying i can't solve this, but it looks like a lot of work and if there's a better approach i'd love some advice.
Here's my sketch of the situation. Apologies if i mis-use terminology!
Speaking of terminology, is there a word for this point ? The point of reflection ? The point of incidence ?
geometry reflection
geometry reflection
asked Jan 29 at 17:09
orion elenzilorion elenzil
1153
asked Jan 29 at 17:09
orion elenzilorion elenzil
1153
asked Jan 29 at 17:09
orion elenzilorion elenzil
1153
asked Jan 29 at 17:09
orion elenzilorion elenzil
1153
asked Jan 29 at 17:09
orion elenzilorion elenzil
1153
1153
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $S’$ be the point on $overline{M_1 M_2}$ that realizes the minimal distance to $S$. Assuming the mirror is vast enough, the line segment $overline{SS’}$ will be the altitude from $S$ to the opposite side in the triangle $SM_1 M_2$. Similarly, let $P'$ be the point on $overline{M_1 M_2}$ that realizes the minimal distance to $P$. Then triangle $SS’Q$ is similar to $PP’R$ because $angle SS'R = angle PP’R = frac{pi}{2}$ and $angle SRS’ = angle PRP'$. If one knows the coordinates of $S$ and $P$ and an equation for the line through $M_1$ and $M_2$, one can easily compute coordinates for $S’$ and $P’$.
By similarity,
$$
frac{S’R}{SS’} = frac{RP’}{PP’} = frac{S’P’ - S’R}{PP’}.
$$
Rearranging, we see that the distance from $S’$ to $R$ along the line of the mirror is given by
$$
S’R = frac{(SS’)(S’P’)}{PP’ + SS’}.
$$
answered Jan 29 at 17:25
Jordan GreenJordan Green
1,146410
$endgroup$
$begingroup$
this is beautiful. thank you so much.
$endgroup$
– orion elenzil
Jan 29 at 18:38
add a comment |
$begingroup$
Construct point $P'$, the reflection of $P$ about the plane of the mirror. The reflection point you seek is then the intersection between line $SP'$ and the plane of the mirror.
answered Jan 29 at 18:17
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
so simple! thank you. these are all such great answers. i'm accepting @jordan green's, as it avoids the further computational step of line intersection.
$endgroup$
– orion elenzil
Jan 29 at 18:37
add a comment |
$begingroup$
Drop perpendiculars from $S$ and $P$ to the mirror, meeting that line at $E$ and $F$. If the desired point is (as you have labeled it) $R$, then $triangle SERsim triangle PFR$ since they are right triangles sharing a common angle at $R$. Thus $frac{ET}{FT} = frac{SE}{PF}$.
answered Jan 29 at 17:20
rogerlrogerl
18k22848
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $S’$ be the point on $overline{M_1 M_2}$ that realizes the minimal distance to $S$. Assuming the mirror is vast enough, the line segment $overline{SS’}$ will be the altitude from $S$ to the opposite side in the triangle $SM_1 M_2$. Similarly, let $P'$ be the point on $overline{M_1 M_2}$ that realizes the minimal distance to $P$. Then triangle $SS’Q$ is similar to $PP’R$ because $angle SS'R = angle PP’R = frac{pi}{2}$ and $angle SRS’ = angle PRP'$. If one knows the coordinates of $S$ and $P$ and an equation for the line through $M_1$ and $M_2$, one can easily compute coordinates for $S’$ and $P’$.
By similarity,
$$
frac{S’R}{SS’} = frac{RP’}{PP’} = frac{S’P’ - S’R}{PP’}.
$$
Rearranging, we see that the distance from $S’$ to $R$ along the line of the mirror is given by
$$
S’R = frac{(SS’)(S’P’)}{PP’ + SS’}.
$$
answered Jan 29 at 17:25
Jordan GreenJordan Green
1,146410
$endgroup$
$begingroup$
this is beautiful. thank you so much.
$endgroup$
– orion elenzil
Jan 29 at 18:38
add a comment |
$begingroup$
Let $S’$ be the point on $overline{M_1 M_2}$ that realizes the minimal distance to $S$. Assuming the mirror is vast enough, the line segment $overline{SS’}$ will be the altitude from $S$ to the opposite side in the triangle $SM_1 M_2$. Similarly, let $P'$ be the point on $overline{M_1 M_2}$ that realizes the minimal distance to $P$. Then triangle $SS’Q$ is similar to $PP’R$ because $angle SS'R = angle PP’R = frac{pi}{2}$ and $angle SRS’ = angle PRP'$. If one knows the coordinates of $S$ and $P$ and an equation for the line through $M_1$ and $M_2$, one can easily compute coordinates for $S’$ and $P’$.
By similarity,
$$
frac{S’R}{SS’} = frac{RP’}{PP’} = frac{S’P’ - S’R}{PP’}.
$$
Rearranging, we see that the distance from $S’$ to $R$ along the line of the mirror is given by
$$
S’R = frac{(SS’)(S’P’)}{PP’ + SS’}.
$$
answered Jan 29 at 17:25
Jordan GreenJordan Green
1,146410
$endgroup$
$begingroup$
this is beautiful. thank you so much.
$endgroup$
– orion elenzil
Jan 29 at 18:38
add a comment |
$begingroup$
Let $S’$ be the point on $overline{M_1 M_2}$ that realizes the minimal distance to $S$. Assuming the mirror is vast enough, the line segment $overline{SS’}$ will be the altitude from $S$ to the opposite side in the triangle $SM_1 M_2$. Similarly, let $P'$ be the point on $overline{M_1 M_2}$ that realizes the minimal distance to $P$. Then triangle $SS’Q$ is similar to $PP’R$ because $angle SS'R = angle PP’R = frac{pi}{2}$ and $angle SRS’ = angle PRP'$. If one knows the coordinates of $S$ and $P$ and an equation for the line through $M_1$ and $M_2$, one can easily compute coordinates for $S’$ and $P’$.
By similarity,
$$
frac{S’R}{SS’} = frac{RP’}{PP’} = frac{S’P’ - S’R}{PP’}.
$$
Rearranging, we see that the distance from $S’$ to $R$ along the line of the mirror is given by
$$
S’R = frac{(SS’)(S’P’)}{PP’ + SS’}.
$$
answered Jan 29 at 17:25
Jordan GreenJordan Green
1,146410
$endgroup$
Let $S’$ be the point on $overline{M_1 M_2}$ that realizes the minimal distance to $S$. Assuming the mirror is vast enough, the line segment $overline{SS’}$ will be the altitude from $S$ to the opposite side in the triangle $SM_1 M_2$. Similarly, let $P'$ be the point on $overline{M_1 M_2}$ that realizes the minimal distance to $P$. Then triangle $SS’Q$ is similar to $PP’R$ because $angle SS'R = angle PP’R = frac{pi}{2}$ and $angle SRS’ = angle PRP'$. If one knows the coordinates of $S$ and $P$ and an equation for the line through $M_1$ and $M_2$, one can easily compute coordinates for $S’$ and $P’$.
By similarity,
$$
frac{S’R}{SS’} = frac{RP’}{PP’} = frac{S’P’ - S’R}{PP’}.
$$
Rearranging, we see that the distance from $S’$ to $R$ along the line of the mirror is given by
$$
S’R = frac{(SS’)(S’P’)}{PP’ + SS’}.
$$
answered Jan 29 at 17:25
Jordan GreenJordan Green
1,146410
edited Jan 29 at 22:58
answered Jan 29 at 17:25
Jordan GreenJordan Green
1,146410
answered Jan 29 at 17:25
Jordan GreenJordan Green
1,146410
answered Jan 29 at 17:25
Jordan GreenJordan Green
1,146410
1,146410
$begingroup$
this is beautiful. thank you so much.
$endgroup$
– orion elenzil
Jan 29 at 18:38
add a comment |
$begingroup$
this is beautiful. thank you so much.
$endgroup$
– orion elenzil
Jan 29 at 18:38
$begingroup$
this is beautiful. thank you so much.
$endgroup$
– orion elenzil
Jan 29 at 18:38
$begingroup$
this is beautiful. thank you so much.
$endgroup$
– orion elenzil
Jan 29 at 18:38
add a comment |
$begingroup$
Construct point $P'$, the reflection of $P$ about the plane of the mirror. The reflection point you seek is then the intersection between line $SP'$ and the plane of the mirror.
answered Jan 29 at 18:17
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
so simple! thank you. these are all such great answers. i'm accepting @jordan green's, as it avoids the further computational step of line intersection.
$endgroup$
– orion elenzil
Jan 29 at 18:37
add a comment |
$begingroup$
Construct point $P'$, the reflection of $P$ about the plane of the mirror. The reflection point you seek is then the intersection between line $SP'$ and the plane of the mirror.
answered Jan 29 at 18:17
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
so simple! thank you. these are all such great answers. i'm accepting @jordan green's, as it avoids the further computational step of line intersection.
$endgroup$
– orion elenzil
Jan 29 at 18:37
add a comment |
$begingroup$
Construct point $P'$, the reflection of $P$ about the plane of the mirror. The reflection point you seek is then the intersection between line $SP'$ and the plane of the mirror.
answered Jan 29 at 18:17
AretinoAretino
25.8k31545
$endgroup$
Construct point $P'$, the reflection of $P$ about the plane of the mirror. The reflection point you seek is then the intersection between line $SP'$ and the plane of the mirror.
answered Jan 29 at 18:17
AretinoAretino
25.8k31545
answered Jan 29 at 18:17
AretinoAretino
25.8k31545
answered Jan 29 at 18:17
AretinoAretino
25.8k31545
answered Jan 29 at 18:17
AretinoAretino
25.8k31545
25.8k31545
$begingroup$
so simple! thank you. these are all such great answers. i'm accepting @jordan green's, as it avoids the further computational step of line intersection.
$endgroup$
– orion elenzil
Jan 29 at 18:37
add a comment |
$begingroup$
so simple! thank you. these are all such great answers. i'm accepting @jordan green's, as it avoids the further computational step of line intersection.
$endgroup$
– orion elenzil
Jan 29 at 18:37
$begingroup$
so simple! thank you. these are all such great answers. i'm accepting @jordan green's, as it avoids the further computational step of line intersection.
$endgroup$
– orion elenzil
Jan 29 at 18:37
$begingroup$
so simple! thank you. these are all such great answers. i'm accepting @jordan green's, as it avoids the further computational step of line intersection.
$endgroup$
– orion elenzil
Jan 29 at 18:37
add a comment |
$begingroup$
Drop perpendiculars from $S$ and $P$ to the mirror, meeting that line at $E$ and $F$. If the desired point is (as you have labeled it) $R$, then $triangle SERsim triangle PFR$ since they are right triangles sharing a common angle at $R$. Thus $frac{ET}{FT} = frac{SE}{PF}$.
answered Jan 29 at 17:20
rogerlrogerl
18k22848
$endgroup$
add a comment |
$begingroup$
Drop perpendiculars from $S$ and $P$ to the mirror, meeting that line at $E$ and $F$. If the desired point is (as you have labeled it) $R$, then $triangle SERsim triangle PFR$ since they are right triangles sharing a common angle at $R$. Thus $frac{ET}{FT} = frac{SE}{PF}$.
answered Jan 29 at 17:20
rogerlrogerl
18k22848
$endgroup$
add a comment |
$begingroup$
Drop perpendiculars from $S$ and $P$ to the mirror, meeting that line at $E$ and $F$. If the desired point is (as you have labeled it) $R$, then $triangle SERsim triangle PFR$ since they are right triangles sharing a common angle at $R$. Thus $frac{ET}{FT} = frac{SE}{PF}$.
answered Jan 29 at 17:20
rogerlrogerl
18k22848
$endgroup$
Drop perpendiculars from $S$ and $P$ to the mirror, meeting that line at $E$ and $F$. If the desired point is (as you have labeled it) $R$, then $triangle SERsim triangle PFR$ since they are right triangles sharing a common angle at $R$. Thus $frac{ET}{FT} = frac{SE}{PF}$.
answered Jan 29 at 17:20
rogerlrogerl
18k22848
answered Jan 29 at 17:20
rogerlrogerl
18k22848
answered Jan 29 at 17:20
rogerlrogerl
18k22848
answered Jan 29 at 17:20
rogerlrogerl
18k22848
18k22848
add a comment |
add a comment |
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