Problem 93 - University Physics with Modern Physics

Question

A convex spherical mirror with a focal length of magnitude 24.0 cm is placed 20.0 cm to the left of a plane mirror. An object 0.250 cm tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object. Where are the two images of the object formed by the spherical mirror that are closest to the spherical mirror, and how tall is each image?

Step-by-Step Solution

Answer

The two images are located at -7.06 cm and -15.48 cm from the mirror, with heights 0.1765 cm and 0.08225 cm respectively.

Step 1: Identify the Setup

We have two mirrors: a convex spherical mirror with a focal length \(|f| = 24.0\) cm, and a plane mirror. The distance between the convex mirror and the plane mirror is 20.0 cm. The object is placed midway, i.e., 10.0 cm from each mirror.

Step 2: Use the Mirror Equation for Convex Mirror

The mirror equation is \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). For a convex mirror, the focal length \(f\) is negative, so \(f = -24.0\) cm.The object distance \(d_o\) from the convex mirror is 10.0 cm.

Step 3: Calculate First Image by Convex Mirror

Substitute \(f = -24.0\) cm and \(d_o = 10.0\) cm into the mirror equation:\[ \frac{1}{-24} = \frac{1}{10} + \frac{1}{d_{i1}} \]Solving gives \(d_{i1} = -7.0588\) cm (negative indicating a virtual image behind the mirror).

Step 4: Calculate Image Height for First Image

The magnification \(m\) is given by \(-\frac{d_i}{d_o}\). For the first image:\[m_1 = -\frac{-7.0588}{10} = 0.7059\]The image height \(h_i = m \times h_o\), so:\[h_{i1} = 0.7059 \times 0.250 = 0.1765\] cm.

Step 5: Calculate Second Image by Plane Mirror

The first virtual image by the convex mirror acts as an object for the plane mirror. This image's distance from the plane mirror is \(20.0 + 7.0588 = 27.0588\) cm on the left side, which forms a second image 27.0588 cm behind the plane mirror.

Step 6: Reflected Image by Convex Mirror

The reflected image by the plane mirror acts as another object for the convex mirror at a distance of \(20 + 27.0588 = 47.0588\) cm from it. Use the mirror equation again:\[\frac{1}{-24} = \frac{1}{47.0588} + \frac{1}{d_{i2}}\]Solving gives \(d_{i2} = -15.4839\) cm.

Step 7: Calculate Image Height for Second Image

The magnification \(m\) for the second image is:\[m_2 = -\frac{-15.4839}{47.0588} = 0.329 \] Thus, \(h_{i2} = 0.329 \times 0.250 = 0.08225\) cm.

Key Concepts

Convex Mirror

A convex mirror, sometimes called a diverging mirror, has a surface that curves outward. Unlike concave mirrors, which can focus light, convex mirrors do the opposite by spreading light out. This makes them valuable for providing a wide field of view. You often see them used in places like parking garages or stores for security.

• Convex mirrors always form virtual images because reflected light from a convex mirror diverges.
• The images are usually upright and smaller than the actual object.

This is key in the exercise, where the convex mirror's properties help form the images of the object in unique ways.

Mirror Equation

The mirror equation is crucial when dealing with curved mirrors. This algebraic relationship helps us determine where images will form when objects are placed in front of mirrors. The equation is:

• \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)

where
- \( f \) is the focal length of the mirror (negative for convex mirrors),
- \( d_o \) is the distance of the object from the mirror, and
- \( d_i \) is the image distance from the mirror.

The sign of each term tells us about the image's nature: real or virtual, upright or inverted. In the example problem, we used this equation twice. First, to find out where the first image forms relative to the convex mirror, and again for the reflected image from the plane mirror.

Image Magnification

Image magnification tells us how much bigger or smaller an image is compared to the actual object. For mirrors, it's represented by:

• \( m = -\frac{d_i}{d_o} \)

Here,
- \( m \) is magnification,
- \( d_i \) is the image distance, and
- \( d_o \) is the object distance.

Negative magnification implies that the image is inverted, though in a convex mirror scenario, the image will still be upright due to the virtual nature. This was used in the example to calculate how tall the images are that the convex mirror formed, revealing that both images are smaller than the object placed between the mirrors.

Virtual Image

Virtual images are formed when light rays appear to diverge from a point behind the mirror. Unlike real images, which can be projected on a screen, virtual images cannot because the light doesn't actually converge. They have the unique trait of appearing inside the mirror.

• Always upright
  
• Found behind the mirror in the mirror's operational space

In the given exercise, the convex mirror creates virtual images. These images seem to be on the opposite side of the mirror from where the object actually is, reflecting the divergent nature of the rays. The image location is also reflected as a negative distance in calculations, marking its virtuality.

Plane Mirror

A plane mirror has a flat reflective surface. They're the kind of mirrors you find in bathrooms or on walls. With plane mirrors, an image is produced through a simpler reflection process:

- The distance between the object and the mirror equals the distance between the image and the mirror. - Plane mirrors always produce virtual images, which appear reversed horizontally.

In the problem, the image created by the convex mirror interacts with the plane mirror. It enables the formation of another image, which, interestingly, acts as a new object for further reflection by the convex mirror. This interplay between the types of mirrors helps in understanding complex image formations.