Problem 45 - University Physics with Modern Physics

Question

You have a spherical mirror with a radius of curvature of \(+20.0 \mathrm{~cm}\) (so it is concave facing you). You are looking at an object whose size you want to double in the image, so you can see it better. Where should you put the object? Where will the image be, and will it be real or virtual?

Step-by-Step Solution

Answer

Where is the image located, and is it real or virtual? Answer: The object should be placed at a distance of 20 cm from the concave mirror to double the size of its image. The image is located at -40 cm from the mirror, indicating that it is a virtual image.

Step 1: Recall the mirror and magnification equations

The mirror equation for spherical mirrors is: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] Where: - \(f\) is the focal length, - \(d_o\) is the object distance (distance from the object to the mirror), and - \(d_i\) is the image distance (distance from the image to the mirror). The magnification equation is: \[ M = -\frac{d_i}{d_o} \] Where: - \(M\) is the magnification factor. Since we want the image to be twice the size of the object, we set \(M = 2\).

Step 2: Calculate the focal length

The radius of curvature \(R\) is given as +20.0 cm. Recall that for concave mirrors, the focal length is half the radius of curvature. Therefore: \[ f = \frac{R}{2} = \frac{20}{2} = 10 \mathrm{~cm} \]

Step 3: Apply the magnification equation M = 2 and solve for \(d_i\) in terms of \(d_o\)

Substitute M = 2 into the magnification equation and solve for \(d_i\): \[ 2 = -\frac{d_i}{d_o} \] \[ d_i = -2d_o \]

Step 4: Substitute \(d_i\) into the mirror equation and solve for \(d_o\)

Replace \(d_i\) with \(-2d_o\) in the mirror equation: \[ \frac{1}{10} = \frac{1}{d_o} - \frac{1}{2d_o} \] Now, solve for \(d_o\): \[ \frac{1}{10} = \frac{1 - 1/2}{d_o} = \frac{1/2}{d_o} \] \[ d_o = 2 \times 10 = 20 \mathrm{~cm} \]

Step 5: Use \(d_o\) to find the image distance \(d_i\)

Now, substitute \(d_o = 20\) cm back into the expression for \(d_i\) from Step 3: \[ d_i = -2d_o = -2(20) = -40 \mathrm{~cm} \]

Step 6: Determine if the image is real or virtual

If the image distance \(d_i\) is positive, the image is real. If the image distance \(d_i\) is negative, the image is virtual. In this case, \(d_i = -40\) cm, so the image is virtual. #Conclusion#: To double the size of the image, place the object at a distance of 20 cm from the concave mirror. The image will be located at -40 cm from the mirror, indicating that it is a virtual image.

Key Concepts

Mirror Equation

The mirror equation is a fundamental principle for understanding how images are formed by spherical mirrors. The equation is expressed as:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]This formula connects three essential variables:

• Focal Length (\(f\)): The distance from the mirror's surface to its focal point.
  
• Object Distance (\(d_o\)): The distance from the object to the mirror.
  
• Image Distance (\(d_i\)): The distance from the image to the mirror.
  

Understanding this equation allows one to predict where an image will form and whether it will be real or virtual based on the position of the object in relation to the mirror. In our specific scenario, solving the mirror equation helps us determine the precise distance at which the object should be placed to achieve the desired image size and location.

Magnification Equation

The magnification equation is crucial for determining the change in size between the object and its image. This equation is written as:
\[ M = -\frac{d_i}{d_o} \]Here, \(M\) represents the magnification factor, indicating how many times larger or smaller the image is compared to the object.

• A positive magnification implies an upright image, while a negative value indicates an inverted image.
  
• In our exercise, we set \(M = 2\), symbolizing the need for the image to be twice the size of the object.
  

When applied, this formula allows us to determine the image distance \(d_i\) in relation to object distance \(d_o\), thus helping in finding the precise positioning needed for the desired magnification.

Virtual Image

A virtual image is a type of image formed when light rays do not really meet but appear to meet when extended backwards. Virtual images have certain characteristics that distinguish them from real images:

• They cannot be projected on a screen because the rays do not converge.
  
• The location of a virtual image is denoted by a negative image distance \(d_i\).
  
• Such images appear behind the mirror or lens systems through which they are viewed.
  

In the given scenario, the image produced is virtual, as indicated by the negative image distance of \(-40\) cm. This means the image is not formed in reality but appears on the same side as the object, making it seem as though one is looking through the mirror.

Concave Mirror

Concave mirrors are mirrors that curve inward, much like the inside of a spoon. They have several unique properties:

• They can produce both real and virtual images depending on the position of the object.
  
• For a concave mirror, the focal length is defined as half of the radius of curvature:
  \[ f = \frac{R}{2} \]
• In our example, with a radius of curvature of \(20\) cm, the focal length \(f\) is \(10\) cm.
  

Concave mirrors are commonly used in devices such as telescopes, flashlights, and makeup mirrors, as they can magnify and converge light to form clear images. In this problem, the concave mirror's capability to form a virtual image through specific object placement was crucial in achieving the desired image magnification.