Problem 103 - University Physics with Modern Physics

Question

You have a camera with a 35.0 - \(\mathrm{mm}\) -focal-length lens and 36.0 -mm- wide film. You wish to take a picture of a \(12.0\)-m-long sailboat but find that the image of the boat fills only \(\frac{1}{4}\) of the width of the film. (a) How fare you from the boat? (b) How much closer must the boat be to you for its image to fill the width of the film?

Step-by-Step Solution

Answer

(a) Approximately 4667 m away; (b) The boat should be around 116.7 m closer.

Step 1: Understand the Problem

You are given a 35.0 mm focal-length lens and a film that is 36.0 mm wide. The image of a 12.0 m sailboat fills only 1/4 of the film's width. We need to find out how far the camera is from the boat and how much closer the boat must be to fill the entire film width.

Step 2: Use the Lens Formula and Scale

The lens formula is \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. The magnification, \( m \), is given by \( \frac{v}{u} = \frac{h_i}{h_o} \), where \( h_i \) is the image height and \( h_o \) is the object height (12.0 m). Since the image fills 1/4 of the film, \( h_i = \frac{36}{4} = 9 \) mm.

Step 3: Calculate Object Distance

Using magnification \( m = \frac{h_i}{h_o} = \frac{9}{12000} \), and knowing \( m = \frac{v}{u} \), we find \( m = \frac{-v}{u} \). The minus sign indicates the image is inverted. \( u = \frac{-h_o \, v}{h_i} = \frac{-12000 \, v}{9} \). Plugging \( v = 35 \) mm (focal length) into the lens formula, solve for \( u \).

Step 4: Determine Image Distance for Full Film Width

If the image fills the full 36.0 mm width, \( h_i = 36 \) mm. Calculate the new magnification \( m = \frac{36}{12000} \) and new object distance using \( u = \frac{-12000 \, v}{36} \) with \( v = 35 \) mm.

Step 5: Find the Difference in Distances

Calculate the difference between the distances found in Step 3 for partial image fill and in Step 4 for full width fill to determine how much closer the boat should be.

Key Concepts

Focal Length

The focal length of a lens is a crucial concept in optics. It determines how strongly a lens focuses or defocuses light. In this problem, the camera lens has a focal length of 35.0 mm. This means that parallel rays of light coming into the lens will converge or diverge to this measurement point behind the lens. It's a fixed property of the lens and helps in forming clear images of objects at different distances.

• A shorter focal length means a wider field of view, while a longer focal length allows you to focus on more distant objects.
• The focal point is where the light rays meet, and from this, the image distance and object distance can be calculated using the lens formula.

Understanding the focal length helps you decide how to adjust the camera to change the size or focus of the image on the film.

Magnification

Magnification tells you how much bigger or smaller the image is compared to the actual object. In this scenario, we use the formula for magnification: \( m = \frac{v}{u} = \frac{h_i}{h_o} \).

• Here, \( m \) is the magnification, \( v \) is the image distance, and \( u \) is the object distance.
• \( h_i \) is the height of the image formed on the film, while \( h_o \) represents the height of the actual object, which is the boat in this case.

If the magnification is less than 1, the image is smaller than the object. In this problem, the partial image is only a quarter of the full film width. This plays into how much of the object can actually be captured relative to its size and the film dimensions. By adjusting magnification, one can change the image's size to fully fit the film width.

Object Distance

Object distance is the distance from the lens to the object being photographed. Using the lens formula \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), where \( u \) is the object distance, we can solve for this distance, given specific conditions like focal length.

• If the object is far from the lens, \( u \) will be large, making it crucial to adjust for clear image formation.
• In Step 3 of the solution, this distance was calculated by incorporating the known image distance \( v \), which corresponds to the film width of only a quarter fill.

Understanding object distance is important as it directly affects how an image appears on the film and helps photographers decide how to position their camera for the desired shot.

Image Distance

Image distance describes where the image will form inside the camera based on how far the object is from the lens. Calculated using the lens formula with given values of focal length and the partial film width used initially.

• This distance, \( v \), is integral to determining the magnification, as it ties directly with the image height and desired image size on the film.
• Adjusting image distance helps in filling either a fraction or the full length of the film, as needed for different photographic goals.
• Calculating it allows photographers to achieve desired results by knowing where the image will be captured with respect to the film.

In optical terms, controlling the image distance is key for capturing precise details of the object, with the correct proportion and positioning within photographs.