New didactic formulation of the laws of reflection of light

Pérez-Rodríguez, Ángel Luis; Martínez-Borreguero, Guadalupe; Suero-López, María Isabel

doi:10.1590/1806-9126-RBEF-2016-0272

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Physics Education Research • Rev. Bras. Ensino Fís. 39 (2) • 2017 • https://doi.org/10.1590/1806-9126-RBEF-2016-0272 copy

New didactic formulation of the laws of reflection of light

Nova formulação didática das leis da reflexão da luz

Authorship SCIMAGO INSTITUTIONS RANKINGS

This work presents a new formulation of the laws of reflection of light, and it is shown the equivalence between the new and the classic formulations. The proposed formulation has a significant educational value, as it allows drawing analogies between the phenomena of light reflection and elastic collisions, which are very well known by students. The new developed statement of the law of reflection allows the resolution of traditional problems of geometrical optics, and in some cases the solution is much simpler and more intuitive than that obtained using the traditional statement of the laws of reflection. As an example, this formulation is applied for the resolution of the classic reflection problem of the three plane mirrors forming a trirectangular trihedron, also known as corner reflector.

Keywords:
Reflection of light; laws of reflection; new formulation; vectors; education

For the first reflection:		For the second reflection:		For the third reflection:
We can state $u_{i}$ , u’, and $N_{1}$ with their direction cosines: $u_{1} : (α_{1}, β_{1} {, γ}_{1})$ $u^{'} : (α_{2}, β_{2} {, γ}_{2})$ $N_{1} : (1,0,0)$		Vectors u’, u”, and $N_{2}$ can be defined by their direction cosines: $u^{'} : (α_{2}, β_{2} {, γ}_{2})$ $u^{″} : (α_{3}, β_{3} {, γ}_{3})$ $N_{2} : (0,1,0)$		Analogously, vectors u”, u $_{r}$ , and $N_{3}$ are defined by their direction cosines: $u^{″} : (α_{3}, β_{3} {, γ}_{3})$ $u_{r} : (α_{4}, β_{4} {, γ}_{4})$ $N_{3} : (0,0,1)$
It is verified that: $u^{'} − u_{1} = 2 N_{1} cos θ$		It is verified that: $u^{″} − u^{'} = 2 N_{2} \cos θ^{'}$		It is verified that: $u_{r} − u^{″} = 2 N_{3} \cos θ^{″}$
Which can be expressed in their components: $α_{2} − α_{1} = 2 \cos θ$ $β_{2} − β_{1} = 0$ $γ_{2} − γ_{1} = 0$		Which again can be expressed in their components: $α_{3} − α_{2} = 0$ $β_{3} − β_{2} = 2 \cos θ^{'}$ $γ_{3} − γ_{2} = 0$		Expressed in their components: $α_{4} − α_{3} = 0$ $β_{4} − β_{3} = 0$ $γ_{4} − γ_{3} = 2 \cos θ^{″}$
$θ$ is the angle between unit vectors −u $_{i}$ and $N_{1}$ then, $\cos θ = − α_{1} ⋅1- β_{1} ⋅0- γ_{1} ⋅0$ and $\cos θ = - α_{1}$ we have then:		$θ$ ’ is the angle between unit vectors −u’ and $N_{2}$ then, $\cos θ^{'} = − α_{2} ⋅0- β_{2} ⋅1- γ_{2} ⋅0$ and $\cos θ^{'} = - β_{2}$ we have then:		$θ$ ” is the angle between unit vectors −u” and $N_{3}$ then, $\cos θ^{″} = − α_{3} ⋅0- β_{3} ⋅0- γ_{3} ⋅1$ and $\cos θ^{″} = - γ_{3}$ we have then:
$α_{2} − α_{1} = − 2 α_{1}$ $β_{2} − β_{1} = 0$ $γ_{2} − γ_{1} = 0$	$α_{2} = − α_{1}$ $β_{2} = β_{1}$ $γ_{2} = γ_{1}$	$α_{3} − α_{2} = 0$ $β_{3} − β_{2} = − 2 β_{2}$ $γ_{3} − γ_{2} = 0$	$α_{3} = α_{2}$ $β_{3} = {− β}_{2}$ $γ_{3} = γ_{2}$	$α_{4} − α_{3} = 0$ $β_{4} − β_{3} = 0$ $γ_{4} − γ_{3} = − 2 γ_{3}$	$α_{4} = α_{3}$ $β_{4} = β_{3}$ $γ_{4} = − γ_{3}$
If we take the relations obtained in the three reflections, we have:
$− α_{1} = α_{2} = α_{3} = α_{4}$			$α_{1} = {− α}_{4}$
$β_{1} = β_{2} = − β_{3} = − β_{4}$			$β_{1} = {− β}_{4}$
$γ_{1} = γ_{2} = γ_{3} = − γ_{4}$			$γ_{1} = {− γ}_{4}$
Therefore, $u_{1} =$ $− u_{r}$ that is, the ray reflected on the third mirror and the incident ray on the first mirror are antiparallel vectors.

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