Problem 10 - College Physics

Question

. Zoom lens, I. A zoom lens is a lens that varies in focal length. The zoom lens on a certain digital camera varies in focal length from 6.50 \(\mathrm{mm}\) to 19.5 \(\mathrm{mm}\) . This camera is focused on an object 2.00 \(\mathrm{m}\) tall that is 1.50 \(\mathrm{m}\) from the camera. Find the distance between the lens and the photo sensors and the height of the image (a) when the zoom is set to 6.50 \(\mathrm{mm}\) focal length and (b) when it is at 19.5 \(\mathrm{mm}\) . (c) Which is the telephoto focal length, 6.50 \(\mathrm{mm}\) or 19.5 \(\mathrm{mm} ?\)

Step-by-Step Solution

Answer

(a) 6.52 mm, 8.69 mm; (b) 19.57 mm, 26.09 mm; (c) 19.5 mm is the telephoto focal length.

Step 1: Understanding the lens formula

The lens formula is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. We will apply this formula to find the image distance \( d_i \) for both given focal lengths.

Step 2: Calculate image distance with 6.50 mm focal length

For (a), set \( f = 6.50 \) mm and \( d_o = 1.50 \) m = 1500 mm. Convert the formula to \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{6.50} - \frac{1}{1500} \). Calculate to find \( d_i \). \[ \frac{1}{d_i} = \frac{1}{6.50} - \frac{1}{1500} \approx 0.15345 \rightarrow d_i \approx 6.52 \text{ mm} \]

Step 3: Calculate height of image with 6.50 mm focal length

The magnification \( m \) is given by \( \frac{h_i}{h_o} = \frac{d_i}{d_o} \) where \( h_i \) is the image height and \( h_o \) is the object height. Rearrange to \( h_i = h_o \cdot \frac{d_i}{d_o} \). Substituting values, \( h_i = 2000 \times \frac{6.52}{1500} \approx 8.69 \text{ mm} \).

Step 4: Calculate image distance with 19.5 mm focal length

Repeat the lens formula for (b) with \( f = 19.5 \) mm. Thus, \( \frac{1}{d_i} = \frac{1}{19.5} - \frac{1}{1500} \approx 0.05114 \rightarrow d_i \approx 19.57 \text{ mm} \).

Step 5: Calculate height of image with 19.5 mm focal length

Using the magnification formula again for (b), \( h_i = 2000 \times \frac{19.57}{1500} \approx 26.09 \text{ mm} \).

Step 6: Determine the telephoto focal length

The telephoto focal length is the longer one, which offers a narrower field of view and larger image. Hence, the telephoto focal length is 19.5 mm.

Key Concepts

Lens Formula

In optics, the lens formula is crucial for understanding how lenses form images. It mathematically relates three key distances in a lens system using the equation: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]- Here, \( f \) is the focal length of the lens.- \( d_o \) represents the distance from the object to the lens.- \( d_i \) is the distance from the lens to the image.
This formula shows how changing one distance affects the others, highlighting the interplay in focusing images.
By rearranging the formula, we can solve for the image distance \( d_i \) using:\[\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}\]This equation is fundamental in tackling problems where you need to determine how far the image is from the lens, which is often the case in practical optical devices like cameras.

Focal Length

The focal length of a lens is a key parameter in optics. It indicates the lens's ability to focus light and form images.

• A smaller focal length results in a wider field of view.
• A larger focal length zooms in more, providing a narrower field, often used in telephoto contexts.

In the context of the exercise, a camera's zoom lens varies its focal length between 6.50 mm and 19.5 mm.
When you hear terms like 'telephoto,' they are usually referring to the longer focal lengths, such as the 19.5 mm in our exercise.
In practical terms, focal length affects how much of a scene is captured in the frame and the level of detail observed in the image.

Magnification

Magnification in optics refers to how much larger or smaller the image appears compared to the object itself. It is given by the formula:\[m = \frac{h_i}{h_o} = \frac{d_i}{d_o}\]- \( h_i \) is the height of the image and \( h_o \) the height of the object.- \( m \) quantifies the change in size.
In the exercise, we computed the image heights for two different focal lengths.
This involves using the image distance \( d_i \) over the object distance \( d_o \), reflecting how much the image size differs from the actual size.

• For 6.50 mm, the image is smaller, with a height of 8.69 mm.
• For 19.5 mm, the image enlarges to 26.09 mm.

This demonstrates how focal length adjustment influences magnification and hence image size.

Image Distance

Image distance \( d_i \) is an outcome of the lens formula and indicates how far the image forms from the lens.
Negative or positive values of \( d_i \) tell us more about the image's formation; whether it's real or virtual.
In our exercise, calculating image distances for different focal lengths (6.50 mm and 19.5 mm) reveals:

• With a focal length of 6.50 mm, the image distance is approximately 6.52 mm.
• With a focal length of 19.5 mm, the image distance extends to around 19.57 mm.

These values help determine the exact placement needed for the camera sensor, crucial for clear photography.
Understanding these settings allows users to adjust lens systems effectively to capture the desired images.