Problem 115 - Fundamentals of Physics

Question

A point object is \(10 \mathrm{~cm}\) away from a plane mirror, and the eye of an observer (with pupil diameter \(5.0 \mathrm{~mm}\) ) is \(20 \mathrm{~cm}\) away. Assuming the eye and the object to be on the same line perpendicular to the mirror surface, find the area of the mirror used in observing the reflection of the point. (Hint: Adapt Fig. \(34-4 .)\)

Step-by-Step Solution

Answer

The area of the mirror used is 0.166 cm wide.

Step 1: Understand the Setup

The object is positioned 10 cm in front of a plane mirror, another 20 cm separates the mirror and the observer's eye. This makes a total distance of 30 cm between the observer's eye and the object, taking the mirror into account.

Step 2: Determine the Geometry

The parallel line (light path) setup forms symmetrical right triangles. The object and its reflection form a triangle with a hypotenuse extending from the object through the mirror to the observer's eye.

Step 3: Calculate the Angle of View

As the observer views the reflection through a pupil with a diameter of 5 mm, we use half the pupil’s diameter for the radius when calculating distances from the center of the pupil, meaning the effective half-width for the mirror viewing is 2.5 mm.

Step 4: Use Similar Triangles

The large triangle (from the object to the eye through the mirror) and small triangle (pupil diameter projection on the mirror) are proportional. Hence, \( \frac{0.25\text{ cm}}{30\text{ cm}} = \frac{x}{10\text{ cm}} \), where \(x\) is the property's width on the mirror.

Step 5: Calculate the Projection Width

Solve for \(x\) using the equation \(x = \frac{0.25 \text{ cm} \times 10 \text{ cm}}{30 \text{ cm}} = \frac{2.5}{30} \approx 0.083\) cm. The effective half-width of the mirror needed is approximately 0.083 cm.

Step 6: Total Area of Mirror

Since the observer views using both edges of the pupil evenly over the linear path, double the half-width (0.083 cm), obtaining the total used mirror width, \(2 \times 0.083\) cm = 0.166 cm.

Key Concepts

Geometric Optics

Geometric optics is a branch of optics that describes light propagation in terms of rays. This is a useful approximation when dealing with the reflection and refraction of light, as it simplifies the complex wave properties of light into more manageable straight lines called rays. In the context of a plane mirror, rays of light from an object are reflected off the mirror at equal angles, following the law of reflection which states that the angle of incidence equals the angle of reflection.

• In this problem, we first consider the setup of a point object placed 10 cm in front of a plane mirror and an observer 20 cm from the mirror on the same line perpendicular to the mirror's surface. This setup allows us to trace the path of light as rays from the object hit the mirror and reflect towards the observer.
• We use geometric optics principles to understand how the rays travel and how they create images through reflection, allowing us to calculate how much of the mirror's surface is actually being used by these rays.

Triangles and Symmetry

The use of triangles and symmetry plays a crucial role in analyzing geometric optics problems. When dealing with mirrors, the concept of symmetry helps us understand how images form and where reflections are directed. Symmetry here means that the paths taken by light to form images have consistent and predictable patterns.
Let's consider the triangles formed in this particular problem. To visualize this, imagine two lines: the direct path from the object to the mirror and the reflected path from the mirror to the observer. These lines, along with the perpendicular from the object to the observer, form two symmetrical right triangles.

• One large triangle consists of the entire path from the object to the observer (via the mirror reflection). This larger view aids in determining the effective angular diameter that limits the light reflecting to the observer.
• The small triangle is the portion of the mirror over which the pupil's projection area is active due to symmetry.

These triangles allow us to utilize geometric relationships, such as those found in similar triangles, to connect the sizes and distances involved in the problem.

Optical Path Calculation

Calculating the optical path in problems involving plane mirrors requires applying principles of similar triangles and proportional reasoning. The aim is to find out how much of the mirror surface is involved in reflecting light into the observer's eye.
Here, the key step is using the large and small triangles formed by the geometric arrangement:

• The observer's pupil has a diameter of 5 mm, suggesting an effective radius of 2.5 mm when imagining how it projects onto the mirror.
• In our exercise, this 0.25 cm represents the half-width of the path where light reflects off the mirror to reach the observer.
• Using the similarity of triangles, we equate the ratios: \( \frac{0.25 \text{ cm}}{30 \text{ cm}} = \frac{x}{10 \text{ cm}} \), giving us the width \( x \) on the mirror.

By solving the equation, we find \( x = 0.083 \text{ cm} \).
Doubling this value gives the total width of the mirror used as 0.166 cm, completing the optical path calculation and providing a comprehensive understanding of how much mirror surface is needed to view the reflected image.