Problem 49 - College Physics

Question

A converging lens has a focal length of 14.0 \(\mathrm{cm} .\) For each of two objects located to the left of the lens, one at a distance of 18.0 \(\mathrm{cm}\) and the other at a distance of \(7.00 \mathrm{cm},\) determine (a) the image position, (b) the magnification, (c) whether the image is real or virtual, and (d) whether the image is erect or inverted. Draw a principal-ray diagram in each case.

Step-by-Step Solution

Answer

Image for first object is real, inverted; image for second object is virtual, erect.

Step 1: Understanding the lens formula

To solve for the image position, we use the lens formula formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance and \( d_i \) is the image distance.

Step 2: Calculating Image Position for First Object

For the first object, where \( d_o = 18.0 \text{ cm} \) and \( f = 14.0 \text{ cm} \), plug these values into the lens formula and solve for \( d_i \): \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{14.0} - \frac{1}{18.0} \). Compute \( d_i \) to find its value.

Step 3: Calculating Image Position for Second Object

For the second object, where \( d_o = 7.00 \text{ cm} \), use the same lens formula. Substitute \( f = 14.0 \text{ cm} \) and \( d_o = 7.00 \text{ cm} \) to find \( d_i \). Compute the result by solving \( \frac{1}{d_i} = \frac{1}{14.0} - \frac{1}{7.00} \).

Step 4: Magnification Calculation

Use the magnification formula: \( M = -\frac{d_i}{d_o} \). For each object, substitute \( d_o \) and the calculated \( d_i \) to find the magnification \( M \).

Step 5: Determining Real or Virtual Image

An image is real if \( d_i \) is positive, and virtual if \( d_i \) is negative. Use the computed \( d_i \) for each object to determine whether the image is real or virtual.

Step 6: Determining Erect or Inverted Image

An image is inverted if the magnification \( M \) is negative and erect if \( M \) is positive. Evaluate the sign of \( M \) for both objects to determine their orientation.

Step 7: Drawing Principal Ray Diagrams

Sketch ray diagrams based on the object distances and focal length. Draw rays to show where the image forms and indicate its nature (real, virtual, erect, inverted) using corresponding paths.

Key Concepts

Focal Length

The focal length of a lens is a measure of how strongly the lens converges or diverges light. For a converging lens, like the one in our exercise, the focal length is positive and represents the distance from the center of the lens to the focal point. When light rays parallel to the principal axis pass through the lens, they converge at this focal point.

• In this exercise, the focal length is given as 14.0 cm.
• The focal length is critical as it helps in determining where the image will form.

By understanding the focal length, you can predict how much the lens will bend the light rays, assisting in calculating image positions.

Lens Formula

The lens formula provides a vital relationship between the object distance, the image distance, and the focal length of a lens. The lens formula is expressed as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where:

• \( f \) is the focal length.
• \( d_o \) is the object distance (distance from the object to the lens).
• \( d_i \) is the image distance (distance from the image to the lens).

To find \( d_i \), rearrange the equation as follows:\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]This allows us to calculate where the image forms relative to the lens whether it’s closer or farther than the object.

Image Magnification

Magnification describes how much larger or smaller the image is compared to the object. It is calculated using the formula:\[ M = -\frac{d_i}{d_o} \]This formula tells us:

• If \( M \) is greater than 1, the image is larger than the object.
• If \( M \) is less than 1, the image is smaller.
• A positive \( M \) indicates that the image is erect (upright).
• A negative \( M \) indicates the image is inverted.

Understanding magnification helps in knowing the scale and orientation of the image.

Real vs. Virtual Image

Understanding whether an image is real or virtual is crucial in optics. Real images are those which can be projected onto a screen:

• They are formed when light actually converges at a point.
• In the lens formula, real images have a positive \( d_i \).
• Real images are often inverted.

In contrast, virtual images cannot be projected and are visualized behind the lens:

• These occur when the outgoing rays appear to diverge from a point.
• Virtual images have a negative \( d_i \) in the lens formula.
• They are generally erect.

By examining the sign of \( d_i \), you can determine the nature of the image, which often defines how the lens system will be utilized in various applications.