Problem 58 - College Physics

Question

.. Where must you place an object in front of a concave mir- ror 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 \( \frac{7R}{10} \) in front of the mirror; the image is \( \frac{7R}{4} \) from the mirror.

Step 1: Understanding the Mirror Equation

The mirror equation is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length of the mirror, \( d_o \) is the object distance, and \( d_i \) is the image distance. For a concave mirror, the focal length \( f \) is given by \( f = \frac{R}{2} \).

Step 2: Understanding Magnification for Mirrors

Magnification \( m \) is given by \( m = -\frac{d_i}{d_o} \). We need the image to be 2.5 times larger and erect, so \( m = 2.5 \). Since the image is erect, the magnification is positive: \( m = +2.5 \).

Step 3: Calculating Image Distance from Magnification

Using \( m = +2.5 \), substitute in the magnification formula: \( 2.5 = \frac{d_i}{d_o} \). Thus, \( d_i = 2.5 d_o \).

Step 4: Substitute in the Mirror Equation

Substituting \( d_i = 2.5 d_o \) into the mirror equation gives: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{2.5 d_o} \].

Step 5: Substituting Focal Length

Since the focal length \( f = \frac{R}{2} \), the mirror equation becomes: \[ \frac{2}{R} = \frac{1}{d_o} + \frac{1}{2.5 d_o} \].

Step 6: Finding Object Distance

Combine the terms: \[ \frac{1}{d_o} + \frac{1}{2.5 d_o} = \frac{2.5 + 1}{2.5d_o} = \frac{3.5}{2.5 d_o} = \frac{2}{R} \]. Solving for \( d_o \), we have: \[ d_o = \frac{3.5R}{2 imes 2.5} = \frac{3.5R}{5} = \frac{7R}{10} \].

Step 7: Calculate Image Position

Now, calculate \( d_i \) using \( d_i = 2.5 d_o \). We have \( d_o = \frac{7R}{10} \), so \( d_i = 2.5 \times \frac{7R}{10} = \frac{17.5R}{10} = \frac{35R}{20} = \frac{7R}{4} \).

Key Concepts

Mirror Equation

The mirror equation is fundamental when studying concave mirrors. It is expressed mathematically as: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Here, \( f \) represents the focal length, \( d_o \) is the object distance, and \( d_i \) the image distance. This equation helps determine where an image forms and its characteristics:

• If \( d_i \) is positive, the image is real and on the same side as the object.
• If \( d_i \) is negative, the image is virtual and located behind the mirror.

Understanding this equation allows us to predict image formation precisely.

Focal Length

For concave mirrors, the focal length \( f \) holds a special relationship with the radius of curvature \( R \). The focal length is simply half of the radius, expressed as:\[ f = \frac{R}{2} \]This characteristic is crucial because:

• It is the point where parallel rays converge after reflecting off the mirror.
• Its positive value indicates the concave mirror’s ability to focus light.

Knowing the focal length is vital for using the mirror equation effectively and predicting image properties.

Magnification

Magnification (\( m \)) indicates how much larger or smaller an image is compared to the object. It is defined as:\[ m = -\frac{d_i}{d_o} \]

• When \( m \) is positive, the image is upright relative to the object.
• When \( m \) is negative, the image is inverted.
• The absolute value of \( m \) greater than 1 means the image is larger than the object.
• If \( m \) is equal to 2.5, as in our problem, it means the image is 2.5 times the size of the object.

With this problem, since \( m \) is positive, it ensures the image is erect.

Object Distance

The object distance \( d_o \) is critical when solving for image properties in relation to the mirror equation. It is the distance from the object to the mirror's surface. In the exercise, we are required to locate the object such that the image is 2.5 times the size of the object and erect:

• Using the magnification and mirror equation, \( d_o \) is calculated to be \( \frac{7R}{10} \).
• This indicates the object's position relative to the mirror's focal point.

Accurate placement of \( d_o \) influences the nature of the image produced.

Image Distance

Image distance \( d_i \) tells us where the image forms relative to the mirror. It is linked to the object distance and magnification:

• For our calculation, we know \( m = 2.5 \), so \( d_i = 2.5 d_o \).
• Substituting our value for \( d_o \) \( \left( \frac{7R}{10} \right) \), \( d_i \) is computed as \( \frac{7R}{4} \).
• This positive value of \( d_i \) indicates a real image formed on the opposite side of the mirror compared to a virtual image.

Through these relationships, we establish where and how the image appears in context to the mirror.