Problem 31 - Fundamentals of Physics

Question

(a) A luminous point is moving at speed \(v_{O}\) toward a spherical 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 \(),(\mathrm{c}) p=8.0 \mathrm{~cm}\) (just outside the focal point), and (d) \(p=10 \mathrm{~mm}\) (very near the mirror).

Step-by-Step Solution

Answer

For (b) \( v_I = -\frac{5}{9} \, cm/s \); (c) \( v_I = -1125 \, cm/s \); (d) \( v_I \approx -100.3 \, cm/s \).

Step 1: Express image and object distances

The spherical mirror formula relates the object distance \( p \), image distance \( q \), and radius of curvature \( r \): \( \frac{1}{p} + \frac{1}{q} = \frac{2}{r} \). Rearranging, we find \( q = \frac{pr}{2p - r} \).

Step 2: Differentiate to find image velocity

To find the velocity of the image \( v_I \), differentiate the image distance \( q \) with respect to time \( t \): \( \frac{dq}{dt} = \frac{d}{dt} \left( \frac{pr}{2p-r} \right) \). Using the quotient rule, we get \( v_I = \frac{r^2 v_O}{(2p - r)^2} \), where \( v_O = \frac{dp}{dt} \). This leads to \( v_I = -\left( \frac{r}{2p-r} \right)^2 v_O \) because the image and object move in opposite directions.

Step 3: Calculate image velocity for \( p = 30 \, cm \)

Substitute \( r = 15 \, cm \), \( v_O = 5.0 \, cm/s \), and \( p = 30 \, cm \) into the derived formula: \( v_I = -\left( \frac{15}{2 \times 30 - 15} \right)^2 \times 5.0 = -\left( \frac{15}{45} \right)^2 \times 5.0 = -\left( \frac{1}{3} \right)^2 \times 5.0 = -\frac{5}{9} \, cm/s \).

Step 4: Calculate image velocity for \( p = 8.0 \, cm \)

Substitute \( p = 8.0 \, cm \) into the formula: \( v_I = -\left( \frac{15}{2 \times 8.0 - 15} \right)^2 \times 5.0 = -\left( \frac{15}{1} \right)^2 \times 5.0 = -225 \times 5.0 = -1125 \, cm/s \).

Step 5: Calculate image velocity for \( p = 10 \, mm \)

Convert \( p = 10 \, mm \) to \( 1 \, cm \) and substitute into the formula: \( v_I = -\left( \frac{15}{2 \times 1 - 15} \right)^2 \times 5.0 = -\left( \frac{15}{-13} \right)^2 \times 5.0 = -\left( \frac{15}{13} \right)^2 \times 5.0 = -\frac{3375}{169} \times 5.0 \approx -100.3 \, cm/s \).

Key Concepts

Understanding Image Velocity in Spherical Mirrors

Imagine a mirror that can reflect images at fascinating speeds! In the world of optics, particularly with spherical mirrors, the **image velocity** plays a key role in determining how fast the reflection of an object is created and changes as the object moves. The image velocity refers to the speed at which the image of an object is moving, relative to the mirror.
Why does this happen? This is due to the change in position of the object which in turn affects its position relative to the mirror. In our particular problem, we can calculate the image velocity by differentiating the formula for image distance, giving us the equation:

• \( v_I = -\left( \frac{r}{2p-r} \right)^2 v_O \)

Here, \(v_O\) is the velocity of the object, \(r\) is the radius of curvature of the mirror, and \(p\) is the object distance. This formula shows that the image velocity not only depends on how fast the object is moving but also on the curvature of the mirror and the object's position.
For students, visualizing this concept may help. Imagine a point of light approaching a concave mirror very fast—it won't just create an image at blink-and-miss speed, but the speed of this image itself will differ depending on how far or close the light is from the mirror.

Decoding Radius of Curvature

Understanding the **radius of curvature** is vital in grasping how curved surfaces, like mirrors, function in optics. The radius of curvature \(r\) represents the distance from the center of the sphere (of which the mirror is a part) to the surface of the mirror.
To picture this, envision a huge circle so massive that its curved line barely seems curved when you see just a section. The radius you measure from the center of this giant circle to its edge is the same concept as the radius of curvature for mirrors and lenses.
The radius of curvature heavily influences the behavior of light and, consequently, the image produced. In our exercise:

• The radius of curvature doubles as the focal length (in spherical mirrors, the focal length \(f\) is given by \(f = \frac{r}{2}\)).
• It dictates how the mirror scatters or converges light rays, affecting the clarity, size, and position of an image.

By controlling this curvature, we can manipulate light to enhance resolution or focus, much like adjusting a camera lens for a crisp picture.

Exploring Concave Mirrors

**Concave mirrors** are segments of a sphere that curve inward, resembling the inside of a spoon. These mirrors are known for converging light rays to a focal point, making them invaluable in several practical applications.
How do concave mirrors work? They utilize the principles of reflection and curvature:

• When parallel light rays hit a concave mirror, they reflect toward a common focal point.
• If an object is placed outside this focal length, the reflected rays converge, producing a real, inverted image. However, when placed within the focal length, the mirror produces a virtual, upright, and magnified image.

In the exercise you are working on, the mirror has a radius of curvature of \(15\) cm. Consequently, this means the focal length is \(7.5\) cm. Understanding where the object stands in relation to this focal length is crucial in predicting how the mirror behaves, especially concerning the speed at which an image forms. Overall, concave mirrors serve special purposes in telescopes, headlights, and even shaving mirrors, wherever you need to focus light to a point.

Applying Optics Formula

The **optics formula** is the underpinning of understanding mirror behavior and image formation. It's a tool that aids in determining relationships between object distance, image distance, and the radius of curvature.
The crucial formula in our scenario is the spherical mirror equation:

• \( \frac{1}{p} + \frac{1}{q} = \frac{2}{r} \)

Here, \(p\) is the object distance, \(q\) is the image distance, and \(r\) is the radius of curvature. By rearranging this formula, students can solve for unknown variables based on given conditions. It's a bit like solving a puzzle where every piece just fits after the calculations.
This formula not only helps in finding where an image forms but extends to calculate things like image velocity as seen in the exercise. Mastery of this formula empowers learners to predict optical behavior, such as how a concave mirror will project an image based on how far an object is placed from it, setting the stage for more advanced studies in optics and its real-world applications.