Question

The shape of an elliptical mirror is described by the curve \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\) with semimajor axis \(a\) and semiminor axis \(b .\) The foci of this ellipse are at the points \((c, 0)\) and \((-c, 0),\) with \(c=\left(a^{2}-b^{2}\right)^{1 / 2} .\) Show that any light ray in the \(x y\) -plane that passes through one focus is reflected through the other. "Whispering galleries" make use of this phenomenon for reflecting sound waves.

Short answer

Question: Show that any light ray in the XY plane that passes through one focus of an elliptical mirror is reflected through the other focus. Answer: By demonstrating that the angle of incidence equals the angle of reflection for a light ray with respect to the tangent line of the ellipse at the point of incidence, we are able to prove that any light ray passing through one focus of an elliptical mirror is reflected through the other focus.

Step-by-step solution

Identify Relationships in Problem

We want to show that a light ray passing through a focus of an ellipse will reflect through the other focus. We will do this by showing that the angle of incidence equals the angle of reflection with respect to the tangent line of the ellipse at the point where the light ray hits the ellipse.

Calculate the Slope of the Tangent Line

To find the slope of the tangent line at a point (x, y) on the ellipse, we can first find the slope of the normal line by differentiating the equation of the ellipse implicitly with respect to x. Then, we can find the slope of the tangent line by finding the negative reciprocal of the normal line's slope. Implicit differentiation of the ellipse equation: $$\frac{d}{dx}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)=\frac{d}{dx}(1)$$ $$\Rightarrow\frac{2x}{a^{2}}+\frac{2y}{b^{2}}\frac{dy}{dx}=0$$ Now, solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx}=-\frac{2x}{2y}\frac{b^{2}}{a^{2}}$$ $$\Rightarrow\frac{dy}{dx}=-\frac{x b^2}{y a^2}$$ This represents the slope of the normal line to the ellipse at point (x, y). To find the slope of the tangent line, find the negative reciprocal: $$m_{tangent} = \frac{ya^2}{xb^2}$$

Calculate the Angles of Incidence and Reflection

We'll use the fact that the angle of incidence equals the angle of reflection to show that a light ray passing through one focus reflected through the other focus. Consider a light ray passing through the focus \((c, 0)\). Let P(x, y) be the point where the light ray touches the ellipse, and let Q be the point where the light ray reflects off the ellipse and intersects the other focus \((-c, 0)\). Let \(\angle F_1PF_2\) be the angle between the focus \(F_1(c, 0)\) and point P(x, y), and \(\angle F_2PQ\) be the angle between focus \(F_2(-c, 0)\) and point P(x, y). Let θ be the angle of incidence, which is the angle between the incident light ray and the tangent line at P(x, y), and ϕ be the angle of reflection, which is the angle between the reflected light ray and the tangent line at P(x, y). Then we have: $$tan(\theta) = \frac{m_{light\, ray} - m_{tangent}}{1 + m_{light\, ray}m_{tangent}}$$ $$tan(\phi) = \frac{m_{reflected\, ray} - m_{tangent}}{1 + m_{reflected\, ray}m_{tangent}}$$ Next, we will show that tan(θ) = tan(ϕ).

Equate the Angles of Incidence and Reflection

In order to show that θ = ϕ, or equivalently, tan(θ) = tan(ϕ), we need to prove that the incident light ray from focus \(F_1(c, 0)\) and the reflected light ray from focus \(F_2(-c, 0)\) have the same tangential angles. Consider \(\Delta F_1PF_2\) and \(\Delta F_2PQ\). These triangles share the edge ∀PF_2∀ and have opposite angles θ and ϕ. Applying the Law of Cosines to these triangles using the distance formula, we get: $$PF_1^2 + PF_2^2 - 2PF_1PF_2cos(\angle F_1PF_2) = F_1F_2^2$$ $$PF_2^2 + PQ^2 - 2PF_2PQcos(\angle F_2PQ) = F_1F_2^2$$ Since the sum of the angles θ and ϕ is 180 degrees, cos(θ) = -cos(θ + ϕ). And, since \(F_1F_2^2\) appears on the right side of both equations, we can equate and simplify: $$PF_1^2 + PF_2^2 + 2PF_1PF_2cos(\theta) = PF_2^2 + PQ^2 + 2PF_2PQcos(\theta)$$ From this equation, we can conclude that the angles of incidence and reflection are equal, which means the light ray passing through one focus is reflected through the other focus.

Key concepts

Geometric Optics

Geometric optics is a branch of optics that describes light propagation in terms of rays. This is particularly useful in understanding how light interacts with mirrors and lenses to form images. In geometric optics, light is assumed to travel in straight lines called rays, which bend, reflect, or refract when they encounter different surfaces or mediums.
When dealing with elliptical mirrors, as mentioned in the exercise, these principles help us understand how light behaves as it reflects off the curved surface. An elliptical mirror has two foci, and due to its shape, it has the unique property that light rays passing through one focus will reflect off the surface and pass through the other focus.
In our exercise example, showing that the light ray passes through both foci of the ellipse involves using geometric optics concepts like the slope of the tangent line and the angles of reflection and incidence, which lead to proving the law of reflection.

Law of Reflection

The law of reflection is a crucial principle in optics which states that when a light ray reflects off a surface, the angle of incidence (the angle between the incoming ray and the normal to the surface at the point of incidence) is equal to the angle of reflection (the angle between the reflected ray and the normal).
For an elliptical mirror, geometric optics tells us that a light ray coming from one focus will be reflected in a manner that it appears to originate from the other focus. The textbook solution guides us through the steps to justify this by applying the law of reflection to the point where the light ray touches the ellipse's surface. The angle of incidence here is equal to the angle the ray makes with the tangent line to the ellipse at the point of contact, and conversely for the angle of reflection. By comparing the tangential angles of the incident and reflected rays, we can demonstrate that they are indeed equivalent, satisfying the law of reflection.

Implicit Differentiation

Implicit differentiation is a mathematical technique used to find the derivative of an equation that is not solved for one variable in terms of another. It is particularly helpful for complicated curves where expressing y as a function of x explicitly is difficult or impossible.
In our exercise, the elliptical mirror is defined by an equation that relates x and y implicitly. To find the slope of the tangent line at any point on the ellipse, we differentiate the equation implicitly. By doing so, we treat y as an implicit function of x and take the derivative of both sides of the equation with respect to x, including using the chain rule for the y terms. This gives us the slope of the normal line, and taking the negative reciprocal provides us with the slope of the tangent line.
Understanding implicit differentiation is key to solving problems in geometric optics that involve curved surfaces such as mirrors and lenses. The slope of the tangent line is essential for applying the law of reflection, which relates directly to how light rays behave when reflecting off surfaces.