Problem 66 - University Physics with Modern Physics

Question

Where must you place an object in front of a concave mirror with radius \(R\) so that the image is erect and \(2\over12\) times the size of the object? Where is the image?

Step-by-Step Solution

Answer

Place the object at \( -\frac{5R}{2} \); the image forms at \( \frac{5R}{12} \).

Step 1: Understanding Mirror and Image Characteristics

A concave mirror can form both real and virtual images. Real images are inverted, while virtual images are erect. To have an erect image, the image must be virtual, meaning the object must be placed within the focal length of the mirror.

Step 2: Formula for Mirror and Magnification

The mirror formula 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. The magnification \( m \) is \( m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \). Given the magnification \( m = \frac{2}{12} = \frac{1}{6} \) and it's positive, we have \( \frac{d_i}{d_o} = \frac{-1}{6} \).

Step 3: Calculating Focal Length

The focal length \( f \) of a concave mirror is half the radius of curvature \( R \), so \( f = \frac{R}{2} \).

Step 4: Express Image Distance in Terms of Object Distance

From the magnification relation \( \frac{d_i}{d_o} = \frac{-1}{6} \), we have \( d_i = -\frac{1}{6}d_o \).

Step 5: Using the Mirror Equation

Substitute \( d_i \) from Step 4 into the mirror equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{(-\frac{1}{6}d_o)} \]. Simplify to find \( d_o \):\[ \frac{1}{f} = \frac{1}{d_o} - \frac{6}{d_o} \] \[ \frac{1}{f} = \frac{-5}{d_o} \].

Step 6: Solve for Object Distance

Rearrange and solve for \( d_o \):\[ d_o = -5f \].Substituting back the value of \( f = \frac{R}{2} \), we get\[ d_o = -5\left(\frac{R}{2}\right) = -\frac{5R}{2} \].

Step 7: Calculate Image Distance

Substitute \( d_o = -\frac{5R}{2} \) into the relation \( d_i = -\frac{1}{6}d_o \) to find \( d_i \):\[ d_i = -\frac{1}{6}\left(-\frac{5R}{2}\right) \] \[ d_i = \frac{5R}{12} \].The image distance is \( \frac{5R}{12} \), positive, indicating a virtual image formed on the same side as the object.

Key Concepts

Mirror Equation

The mirror equation is a fundamental formula used to relate the object distance, image distance, and focal length of a spherical mirror. This equation can be written as: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Here, \( f \) represents the focal length of the mirror, \( d_o \) is the object distance from the mirror, and \( d_i \) is the distance from the mirror to the image.
The mirror equation allows us to understand how an object and its image relate concerning a mirror's curvature. Understanding this relationship is crucial when determining the image's position and type, whether it be real or virtual.

• For concave mirrors, if an image is virtual, the object must be placed within the focal length.
• The equation helps calculate where the image will be formed, whether in front of the mirror or behind, depending on the distances involved.

Magnification

Magnification describes how much larger or smaller an image is than the object itself. In mirror optics, it is defined as:\[ m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \]where \( h_i \) is the image height and \( h_o \) is the object height, \( d_i \) and \( d_o \) are the image and object distances respectively.
Magnification can be positive or negative. A positive magnification indicates an erect (upright) image, while a negative sign indicates an inverted image. In the context of concave mirrors, if an image is erect and magnification is positive as in this problem's solution, it highlights that the image is virtual.

• Erect images imply a virtual image, meaning location is on the same side as the object.
• The magnitude of magnification can help determine how large or small the virtual image appears compared to the object.

Focal Length

The focal length, denoted as \( f \), is the distance between the mirror's surface and the focal point where parallel rays of light converge. For concave mirrors, the focal length is half of the radius of curvature \( R \). Therefore, the relation is expressed as:\[ f = \frac{R}{2} \]
Knowing the focal length is vital because it determines how the mirror will focus incoming light beams. It also plays a key role in the mirror equation as we saw earlier. In practical applications, the focal length allows us to calculate the precise positions of the object and image, ensuring correct predictions of image properties.

• A concave mirror's focal point allows us to focus light, creating sharp images at certain distances.
• Understanding \( f \) is crucial for manipulating real-world imaging systems, such as telescopes and cameras.

Real and Virtual Images

When light reflects off a concave mirror, it can form either real or virtual images. The nature of these images is determined by the object's distance from the mirror compared to the focal length.
Real images are formed when light converges at a point after reflecting off the mirror. These images are inverted and can be projected onto a screen because they are formed on the opposite side of the mirror.
Virtual images, on the other hand, appear to be on the same side as the object. These images are erect (upright), and they cannot be cast on a screen because they are formed by the extension of diverging light rays.

• Real images are inverted and physical, meaning they can be caught on a surface; virtual images cannot.
• Virtual images appear larger with concave mirrors if the object is placed within the focal length, creating a unique magnification effect.

Understanding these differences helps in applications like eyeglasses or mirrors where the orientation and position of the image relative to the object are crucial.