Problem 130 - Fundamentals of Physics

Question

A small cup of green tea is positioned on the central axis of a spherical mirror. The lateral magnification of the cup is \(+0.250\), and the distance between the mirror and its focal point is \(2.00 \mathrm{~cm}\). (a) What is the distance between the mirror and the image it produces? (b) Is the focal length positive or negative? (c) Is the image real or virtual?

Step-by-Step Solution

Answer

(a) Image distance = 1.50 cm. (b) Focal length is positive. (c) The image is virtual.

Step 1: Understand the problem statement

We need to determine three things: (a) the image distance from the mirror, (b) the sign of the focal length, and (c) whether the image is real or virtual. We are given a lateral magnification of +0.250 and a focal distance of 2.00 cm. Let's use these to calculate the required values.

Step 2: Use magnification formula for image distance

The lateral magnification (\(M\)) is given by the formula:\[M = \frac{-i}{o}\]where \(i\) is the image distance and \(o\) is the object distance. Since \(M = 0.250\), we can express the image distance in terms of the object distance:\[-i = 0.250o\]Thus,\[i = -0.250o\]

Step 3: Use mirror equation to find image distance

The mirror equation is:\[\frac{1}{f} = \frac{1}{o} + \frac{1}{i}\]where \(f\) is the focal length, \(o\) is the object distance, and \(i\) is the image distance. Substitute the expression for \(i\) from Step 2:\[\frac{1}{f} = \frac{1}{o} - \frac{1}{0.250o}\]\[\frac{1}{f} = \frac{1}{o} - \frac{4}{o}\]\[\frac{1}{f} = \frac{-3}{o}\]

Step 4: Calculate object distance using focal length

From Step 3, \(\frac{1}{f} = \frac{-3}{o}\), which implies:\[o = -3f\]Substituting \(f = 2.00 \mathrm{~cm}\):\[o = -3 \times 2.00 \mathrm{~cm} = -6.00 \mathrm{~cm}\]

Step 5: Calculate image distance

Substitute \(o = -6.00 \mathrm{~cm}\) into the expression from Step 2:\[i = -0.250 \times (-6.00 \mathrm{~cm}) = 1.50 \mathrm{~cm}\]So, the image distance is 1.50 cm.

Step 6: Evaluate sign of focal length

Since we calculated the object distance as negative and \(f = 2.00 \mathrm{~cm}\), it reinforces that the focal length is positive. Therefore, the mirror is concave.

Step 7: Determine the type of image (real or virtual)

The positive image distance of 1.50 cm implies that the image is virtual, as virtual images are produced when objects are placed within the focal length.

Key Concepts

Lateral Magnification

Lateral magnification is a measure of how much larger or smaller an image is compared to the object. It is represented by the symbol \( M \). For mirrors, the magnification formula is:

• \( M = \frac{-i}{o} \)

where \( i \) is the image distance and \( o \) is the object distance.
In our exercise, the given magnification is \(+0.250\), indicating that the image is smaller than the object. A positive magnification suggests that the image is upright. This is an important clue for identifying whether an image is real or virtual.
When you find lateral magnification, it helps predict the orientation and size of the image compared to the original object.

Mirror Equation

The mirror equation relates the object distance, image distance, and focal length of a spherical mirror. It is written as:

• \( \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \)

This equation is fundamental for solving problems related to both concave and convex mirrors.
In this particular scenario, we used the expression derived from the magnification formula and substituted it into the mirror equation. This substitution helped determine the relationship between the image and the object distance.
Solving the equation helps determine the values needed, like the object distance or the image distance, as it did in the exercise where it revealed the image distance as 1.50 cm.

Virtual Image

A virtual image is formed when the outgoing rays diverge, and the image appears to come from a location behind the mirror. They appear upright and cannot be projected onto a screen.
In the exercise, a positive image distance indicates a virtual image, since the light rays appear to diverge from behind the mirror. This is consistent with the magnification being positive, which further supports the upright position of the image.
Recognizing a virtual image is essential as it informs us about its nature compared to real images, which are inverted and can be caught on a screen.

Focal Length

Focal length is a property of a spherical mirror that defines the distance from the mirror to its focal point (where parallel rays converge or appear to diverge).
Whether the focal length is positive or negative helps determine the type of mirror you're working with.

• A positive focal length indicates a concave mirror.
• A negative focal length indicates a convex mirror.

In the exercise, the focal length is given as +2.00 cm because it was determined by a positive magnification outcome, suggesting a concave mirror. Understanding the sign of the focal length helps in the accurate identification of the mirror type and the nature of forming images.