(a) A luminous point is moving at speed vO toward a spheri-...

(a) A luminous point is moving at speed $v_{O}$ toward a spheri- cal mirror with radius of curvature $r,$ along the central axis of the mirror. Show that the image of this point is moving at speed $$v_{I}=-\left(\frac{r}{2 p-r}\right)^{2} v_{O}$$ where $p$ is the distance of the luminous point from the mirror at any given time. Now assume the mirror is concave, with $r=15 \mathrm{cm}$ ,and let $v_{O}=5.0 \mathrm{cm} / \mathrm{s}$ . Find $v_{I}$ when $(\mathrm{b}) p=30 \mathrm{cm}$ (far outside the focal point), (c) $p=8.0 \mathrm{cm}$ (just outside the focal point), and (d) $p=10 \mathrm{mm}($ very near the mirror).

(a) A luminous point is moving at speed $v_{O}$ toward a spheri- cal mirror with radius of curvature $r,$ along the central axis of the mirror. Show that the image of this point is moving at speed
$$v_{I}=-\left(\frac{r}{2 p-r}\right)^{2} v_{O}$$
where $p$ is the distance of the luminous point from the mirror at
any given time. Now assume the mirror is concave, with $r=15 \mathrm{cm}$ ,and let $v_{O}=5.0 \mathrm{cm} / \mathrm{s}$ . Find $v_{I}$ when $(\mathrm{b}) p=30 \mathrm{cm}$ (far outside the
focal point), (c) $p=8.0 \mathrm{cm}$ (just outside the focal point), and
(d) $p=10 \mathrm{mm}($ very near the mirror).

Key Concepts

Mirror Equation

The mirror equation relates the object distance (p), the image distance (q), and the focal length (f) for spherical mirrors by using the formula 1/p + 1/q = 1/f. This fundamental relationship enables the calculation of where an image will form given any object position and is the basis for understanding how mirrors produce images.

Focal Length

The focal length of a spherical mirror is defined as the distance from the mirror at which parallel light rays converge (or appear to diverge) after reflection. For spherical mirrors, the focal length is directly related to the radius of curvature by the relation f = r/2, a key concept that connects the geometry of the mirror to its optical properties.

Differentiation in Dynamic Imaging

In dynamic imaging problems, where the object is in motion, the position of the image changes over time. By differentiating the mirror equation with respect to time, one can derive a mathematical relationship that connects the rate of change of the object distance to the rate of change of the image distance, thus allowing the calculation of the image speed as a function of the object's speed.

Sign Conventions

Sign conventions in mirror optics specify the positive and negative directions for object distances, image distances, and focal lengths. Correctly applying these conventions is crucial for obtaining accurate results in both static and dynamic analyses of image formation, ensuring that the calculated positions and speeds have the appropriate directions relative to the mirror.

Transcript

00:01 For our question, we're told that the radius of curvature r is equal to 15 centimeters, and v0, the velocity of the object, is equal to 5 centimeters per second.

00:09 And it wants us to find a relationship for the velocity of the image as shown in the question.

00:16 So to do that, we're going to start with our equation, which says that 1 over the distance to the image, i, plus 1 over the distance to the object, p, is equal to 1 over the focal length, f, or 2 divided by the radius of curvature, we're asked to put things in terms of r, so let's go ahead and write the expression this way.

00:36 So we can solve for i, and we see that i is equal to p times r divided by 2 times p minus r over 2.

00:56 We can simplify this a little bit further, and we find that i is equal to p times r divided by 2p minus r.

01:07 I just distributed that 2 all the way through.

01:10 Okay.

01:11 So now let's take the derivative, because the velocity is the derivative with respect to time.

01:16 So, d .i by dt, the derivative of i with respect to t, is equal to the velocity of the image.

01:22 So this is equal to what we're going to call v of i, which is what we're asked to solve for in the expression.

01:28 So this is equal to the derivative, d by dt, of what's on the right side of the equation, which is p, times r divided by 2p minus r well the derivative of r with respect to time is zero because there's no time dependence there it's p that changes with respect to time so taking the derivative of this we find that this is equal to r divided by 2 p minus r squared multiplied by the derivative of what's inside which was d p by d t but d p by d t is the velocity of the object.

02:14 So what we find here is v sub i is equal to r divided by, i had a misprint here.

02:25 This is supposed to be 2p minus r, not 2p times r.

02:28 So this is 2p minus r.

02:33 So r divided by 2p minus r, that whole thing squared, multiplied by d t, which is the velocity of the object, which we called v .0.

02:50 And this is what we were asked to prove for part a, so we can go ahead and box all of that in as our solution for part a.

02:56 Now, we know r and we know the velocity of the object.

03:00 And then for part b, it says, okay, well, now let's consider a situation where the distance p is equal to 30 centimeters...

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