Problem 68 - 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\(\frac{1}{2}\) times the size of the object? Where is the image?

Step-by-Step Solution

Answer

Place the object at \(\frac{7R}{10}\) from the mirror. The image is at \(\frac{7R}{4}\).

Step 1: Understanding Mirror Formula and Magnification

For a concave mirror, the mirror formula is \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance. The magnification \(m\) is given by \( m = -\frac{v}{u} \). An erect image implies that \(m\) is positive, so \(m = \frac{v}{u} \). We are given that \(m = 2.5\).

Step 2: Determine the Focal Length

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

Step 3: Set Up the Given Magnification Condition

We know that \(m = 2.5\), hence \( \frac{v}{u} = 2.5 \). This means that \( v = 2.5u \).

Step 4: Substitute and Solve the Mirror Equation

Substitute \(v = 2.5u\) into the mirror formula: \( \frac{1}{f} = \frac{1}{2.5u} + \frac{1}{u} \). Simplify to find \(u\): \( \frac{1}{f} = \frac{3.5}{2.5u} \), leading to \( u = \frac{3.5f}{2.5} \).

Step 5: Express Object Distance in Terms of Radius of Curvature

Replace \(f\) with \(\frac{R}{2}\) in the found expression for \(u\): \( u = \frac{3.5}{2.5} \cdot \frac{R}{2} = \frac{7R}{10} \). Thus, the object is placed at a distance of \(\frac{7R}{10}\) from the mirror.

Step 6: Calculate Image Distance

Substitute \( u = \frac{7R}{10} \) into \( v = 2.5u \) to find the image distance: \( v = 2.5 \times \frac{7R}{10} = \frac{7R}{4} \). Thus, the image is located at a distance of \(\frac{7R}{4}\) from the mirror.

Key Concepts

Mirror Formula

The mirror formula is essential for understanding how images are formed by mirrors, particularly for concave mirrors. This formula is expressed as: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] where:

• \( f \) is the focal length,
• \( v \) is the image distance,
• \( u \) is the object distance.

A concave mirror allows parallel rays of light to converge, forming images. By applying the mirror formula, one can precisely calculate where an image will appear in relation to the object and the mirror. Understanding this relationship helps predict the nature and position of the image formed.

Magnification

Magnification describes how much larger or smaller an image appears compared to the actual object. It is given by: \[ m = \frac{v}{u} \] For lenses and mirrors:

• If \( m > 1 \), the image is magnified larger than the object.
• If \( m < 1 \), the image is smaller than the object.
• If \( m = 1 \), the image size is equal to the object size.

For erect images formed by concave mirrors, magnification is positive (\( m = \frac{v}{u} \)). This translates to an upright image. In our exercise, a magnification of 2.5 indicates the image is not only upright but also 2.5 times larger than the object.

Image Distance

Image distance refers to the distance between the image and the mirror along the principal axis. This is represented by \( v \) in the mirror formula. Knowing the image distance is crucial for determining where the final image will form relative to the mirror and the object. In the given exercise, using the relationship \( v = 2.5u \), it was determined that the image distance \( v \) is \( \frac{7R}{4} \), which tells us precisely where to look for the image.

Focal Length

The focal length is a vital characteristic of mirrors that determines how they converge or diverge light rays. It is half the radius of curvature, so for a mirror with radius \( R \), the focal length \( f \) is: \[ f = \frac{R}{2} \] A concave mirror focuses light to a point called the focal point, which is located at this distance from the mirror itself. Understanding the focal length is essential when calculating how objects and images relate spatially. In the exercise, this value helped find the correct placements of both object and image along the principal axis.