Problem 40 - University Physics with Modern Physics

Question

An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.00 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

Step-by-Step Solution

Answer

(a) Focal length is -28.8 cm, lens is diverging. (b) Image is 18.0 mm tall, erect.

Step 1: Understanding Lens Formula

To find the focal length, we'll use the lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(d_o\) is the object distance and \(d_i\) is the image distance. Since the object is to the left of the lens, \(d_o = -16.0\, \text{cm}\). The image is to the right, so \(d_i = 36.0\, \text{cm}\).

Step 2: Calculate Focal Length

Plug the values into the lens formula: \(\frac{1}{f} = \frac{1}{-16} + \frac{1}{36}\). Calculating, we get:\[\frac{1}{f} = -\frac{1}{16} + \frac{1}{36} = \frac{-36 + 16}{576} = \frac{-20}{576} = -\frac{5}{144}.\] Thus, \(f = -28.8\, \text{cm}\).

Step 3: Determine Lens Type

Since the focal length is negative, the lens is diverging.

Step 4: Use Magnification Formula

The magnification \(m\) is given by \(m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\). Given \(h_o = 8.00\, \text{mm}\), we calculate \(\frac{h_i}{8} = -\frac{36}{-16}\).

Step 5: Calculate Image Height

Solving for \(h_i\), \(h_i = 8 \times \frac{36}{16} = 8 \times 2.25 = 18.0\, \text{mm}\). The positive magnification indicates the image is erect.

Step 6: Draw Principal-Ray Diagram

For a diverging lens, draw the principal rays. A ray parallel to the axis refracts as if it originated from the focal point on the same side. All rays appear to diverge from that focal point. The image forms upright on the same side as the object.

Key Concepts

Focal Length Calculation

When dealing with lenses, one of the key characteristics we need to determine is the focal length, which is crucial for understanding how a lens affects light. The focal length (\( f \)) can be calculated using the lens formula: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where \( d_o \) is the object distance, and \( d_i \) is the image distance. In our exercise, the object is positioned at \( -16.0 \, \text{cm} \) to the left of the lens, indicating a negative object distance, whereas the image appears \( 36.0 \, \text{cm} \) to the right, allowing us to use a positive image distance. By inputting these values into our formula, we compute the following:

• \( \frac{1}{f} = \frac{1}{-16} + \frac{1}{36} \)
• \( \frac{1}{f} = \frac{-36 + 16}{576} = \frac{-20}{576} \)
• \( f = -28.8 \, \text{cm} \)

The negative focal length confirms the presence of a diverging lens, which diverges light rays outwards.

Image Magnification

Image magnification tells us how much larger or smaller an image is compared to the object. The magnification formula is:\[ m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \]where \( h_i \) is the image height and \( h_o \) is the object height. In this case, the object has a height of \( 8.00 \, \text{mm} \). Using the previously given distances, we find:

• \( m = -\frac{36}{-16} = 2.25 \)

This indicates that the image is \( 2.25 \) times the height of the object. Therefore, \( h_i = 8.00 \, \text{mm} \times 2.25 = 18.0 \, \text{mm} \). The positive value of magnification implies the image is upright.

Diverging Lens

A diverging lens, often termed a concave lens, spreads out light rays that have been refracted through it. This property is evident when the focal length is negative, as calculated in the previous steps. Here’s how a diverging lens operates:

• The parallel rays incident on the lens appear to diverge away and give the impression they are emanating from a single point (the focal point).
• For an object placed on one side of the lens, the image forms upright and on the same side of the lens as the object.

These characteristics allow diverging lenses to be widely used in applications like eyeglasses for nearsightedness, where they help to spread light before it reaches the eyes.

Principal-Ray Diagram

A principal-ray diagram is valuable for visualizing how rays interact with lenses. It aids in predicting where the image will form and its orientation. To draw a principal-ray diagram for a diverging lens:

• Start by drawing the optical axis, then place the lens in the center.
• A ray parallel to the optical axis will refract through the lens as if it came from the focal point on the same side as the object.
• Another ray through the center of the lens will pass without deviation.
• The extensions of the refracted rays will meet on the same side as the object, locating the virtual image there.

This diagram reveals that for diverging lenses, images are smaller, erect, and located on the same side as the object, which is consistent with our earlier calculations.