Problem 13 - Applied Physics

Question

What is the height of an image in a truck mirror when the object is \(1.20 \mathrm{~m}\) tall and is standing \(7.60 \mathrm{~m}\) from the mirror and the image is formed \(1.50 \mathrm{~m}\) from the mirror?

Step-by-Step Solution

Answer

The image height is approximately \(0.237 \mathrm{~m}\) (inverted).

Step 1: Identify Given Information

We are given the height of the object as \(1.20\, \mathrm{m}\), the distance of the object from the mirror as \(7.60\, \mathrm{m}\), and the distance from the mirror at which the image is formed as \(1.50\, \mathrm{m}\).

Step 2: Apply Mirror Formula

Since the distances are given, we use the magnification formula related to mirrors: \( m = \frac{-v}{u} \), where \(m\) is the magnification, \(v\) is the image distance (\(1.50\, \mathrm{m}\)), and \(u\) is the object distance (\(7.60\, \mathrm{m}\)).

Step 3: Calculate Magnification

Plug the given values into the magnification formula: \( m = \frac{-1.50}{7.60} = -0.1974 \). The negative sign denotes the image is inverted with respect to the object.

Step 4: Determine Image Height

Use the magnification definition in terms of height: \( m = \frac{h'}{h} \), where \(h'\) is the image height and \(h\) is the object height (\(1.20\, \mathrm{m}\)). Substitute \(m = -0.1974\) and \(h = 1.20\, \mathrm{m}\): \( -0.1974 = \frac{h'}{1.20} \).

Step 5: Solve for Image Height

Solve for \(h'\): \( h' = 1.20 \times (-0.1974) = -0.23688 \). The height of the image is approximately \(-0.237\, \mathrm{m}\). The negative sign indicates the image is inverted relative to the object.

Key Concepts

Image height calculation

In optical physics, calculating the height of an image in a mirror involves understanding both the distance of the object and the image relative to the mirror. The mirror formula, paired with the magnification formula, helps solve for the image height effectively.

To find the height of an image (denoted as \( h' \)) when given the height of the object and the distances involved, use the formula for magnification: \( m = \frac{h'}{h} \). Here, \( h \) represents the height of the actual object.

This formula rearranges to solve for \( h' \) as follows:
\[ h' = m \times h \] Substitute the values of magnification and object height into this formula to obtain the image height. Be mindful of the sign, as a negative result indicates an inverted image. Using the provided exercise example, the image height calculates to \( -0.237 \) meters, confirming that the image is both smaller and inverted.

Magnification in mirrors

Magnification describes how much larger or smaller an image appears in comparison to the original object. In mirrors, the magnification formula is expressed as \( m = \frac{-v}{u} \), where \( v \) is the image distance, and \( u \) is the object distance.

Key considerations include:

• A positive magnification value signifies an upright image.
• A negative magnification value shows the image is inverted.
• If the magnitude is greater than 1, the image is larger than the object.
• If less than 1, the image is smaller.

In our case, the magnification is \(-0.1974\), indicating an inverted and smaller image. This conclusion stems from the negative sign and the fact that \( |-0.1974| < 1 \). By relating these concepts, the magnification formula helps in predicting both the orientation and size of the resulting image.

Optical physics concepts

Optical physics covers various phenomena involving light and its interaction with mirrors, lenses, and other optical devices. By leveraging core concepts like reflection, refraction, and magnification, we gain a tangible understanding of how images are formed.

Reflection involves the bouncing back of light rays when they encounter a reflective surface, such as a mirror. The distance an image appears from the mirror depends on the curvature type and the object distance. Mirrors can create virtual or real images, which are determined by their specific geometry and the light path involved.

Moreover, refraction, the bending of light as it passes through different media, pairs closely with reflection in various optical devices beyond mirrors. Utilizing the mirror formula and associated magnification equations offers critical insights into the optical physics behind everyday and specialized applications, thereby enabling accurate calculations and predictions of image properties.