Problem 139 - Fundamentals of Physics

Question

A grasshopper hops to a point on the central axis of a spherical mirror. The absolute magnitude of the mirror's focal length is \(40.0 \mathrm{~cm}\), and the lateral magnification of the image produced by the mirror is \(+0.200\). (a) Is the mirror convex or concave? (b) How far from the mirror is the grasshopper?

Step-by-Step Solution

Answer

(a) Convex mirror; (b) 160 cm from the mirror.

Step 1: Understanding Mirror Characteristics

Convex mirrors have positive focal lengths, meaning they diverge light, while concave mirrors have negative focal lengths and converge light. The magnification given is positive, suggesting an upright image, which typically occurs with virtual images in a convex mirror.

Step 2: Determine the Type of Mirror

Given that the lateral magnification is positive, this suggests that the image formed is virtual and upright. In mirrors, a positive magnification implies the object is closer than the focal point in a convex mirror. Thus, we begin by identifying the mirror as convex.

Step 3: Use the Mirror Equation

The mirror equation is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. Since the mirror is convex, \( f = -40.0 \text{ cm} \).

Step 4: Relate Magnification to Distances

The magnification \( m \) is related to the distances by \( m = \frac{-d_i}{d_o} \). Given \( m = +0.200 \), we substitute into this equation: \( +0.200 = \frac{-d_i}{d_o} \).

Step 5: Express Image Distance in Terms of Object Distance

From the magnification equation, we rearrange to find \( d_i = -0.200 \, d_o \). This means the image distance is a fraction of the object distance.

Step 6: Substitute and Solve for Object Distance

Substitute \( d_i = -0.200 \, d_o \) into the mirror equation: \( \frac{1}{-40.0} = \frac{1}{d_o} - \frac{1}{0.200 \, d_o} \). Simplifying gives: \( \frac{1}{-40.0} = \frac{1 - 5}{d_o} = \frac{-4}{d_o} \).

Step 7: Calculate Object Distance

Solve the equation \( \frac{-4}{d_o} = \frac{1}{-40.0} \). Multiply both sides by \( d_o \) and \(-40.0\) to isolate \( d_o \): \( -40.0 \times -4 = d_o \), yielding \( d_o = 160.0 \).

Step 8: Assign Correct Signs and Solve for Object Position

Finally, convert \( d_o \) to cm by retaining the positive sign as the distance is given positively: \( d_o = 160.0 \text{ cm} \).

Key Concepts

Convex Mirror

Convex mirrors are unique because they curve outward, diverging light rays that strike their surface. These mirrors are commonly used in vehicle side mirrors, offering a wider field of view. In terms of optics, convex mirrors always form virtual images. This means the images appear to be behind the mirror even though they cannot be projected on a screen. A virtual image is always upright and reduced, appearing smaller than the actual object.

Since the focal point of a convex mirror is virtual, its focal length carries a negative sign when performing calculations. This is a crucial aspect in distinguishing it from a concave mirror, which behaves differently. The positive magnification given in the problem indicates the image is upright, typical for virtual images in convex mirrors. This is why for this problem, we conclude that the mirror in question is convex.

Concave Mirror

A concave mirror has a reflective surface that curves inward, much like the interior of a bowl. These mirrors are capable of converging light rays to a focal point in front of the mirror. Unlike convex mirrors, concave mirrors can produce both real and virtual images.

When an object is placed beyond the focal point of a concave mirror, it creates a real and inverted image. This image can be captured on a screen. However, if the object is within the focal length, a virtual, upright, and magnified image is produced. For clarification, real images appear smaller and flipped, while virtual images appear larger and upright.

Concave mirrors are often found in applications requiring focused light, such as telescopes, headlamps, and shaving mirrors.

Mirror Equation

The mirror equation is fundamental in solving optics problems related to spherical mirrors, be they convex or concave. It mathematically relates the object distance ( d_o ), image distance ( d_i ), and focal length ( f ). The equation is expressed as:

\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

This equation helps determine where an image will form when an object is placed at a certain distance from a mirror and vice versa. For convex mirrors, the focal length ( f ) is negative, while for concave mirrors, it is positive.

When using this equation, positive and negative signs are important, indicating the direction and nature of image formation (real or virtual). Solving problems using this equation often requires rearranging terms and solving for unknown variables.

Lateral Magnification

Lateral magnification ( m ) describes how the size of an image compares to the object in spherical mirrors. It is given by the ratio of the image height to the object height and also by the ratio of the image distance to the object distance:

\[ m = \frac{h_i}{h_o} = \frac{-d_i}{d_o} \]

In this relation, the negative sign is key. It signifies that real images are inverted. However, when magnification is positive, it indicates an upright and typically virtual image.

The value of m can vary. If m is greater than 1, the image is larger than the object. If less than 1, the image is smaller. Understanding lateral magnification helps us interpret the nature of the image — size, orientation, and type — when observed through a spherical mirror.