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A camera lens has a focal length of 200 \(\mathrm{mm}\) . How far from the lens should the subject for the photo be if the lens is 20.4 \(\mathrm{cm}\) from the film?

Short Answer

Expert verified
The subject should be approximately 1020.4 cm from the lens.

Step by step solution

01

Understanding the Lens Formula

The lens formula is given by \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), where \( f \) is the focal length of the lens, \( v \) is the distance from the lens to the image (film), and \( u \) is the distance from the lens to the object (subject). We need to find \( u \).
02

Converting Units

First, ensure all measurements are in the same units. Convert the distance from the lens to the film from centimeters to millimeters: \( 20.4 \text{ cm} = 204 \text{ mm} \). Now, the given focal length \( f = 200 \text{ mm} \) and distance to the image \( v = 204 \text{ mm} \).
03

Rearranging the Lens Formula

To find \( u \), rearrange the lens formula: \( \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \).
04

Substituting Values into Formula

Substitute the given values into the rearranged formula: \( \frac{1}{u} = \frac{1}{200} - \frac{1}{204} \). This simplifies the problem by allowing us to find \( \frac{1}{u} \).
05

Calculating \( \frac{1}{u} \)

Calculate \( \frac{1}{u} \) by finding the numerical difference: \( \frac{1}{200} = 0.005 \) and \( \frac{1}{204} \approx 0.004902 \). Thus, \( \frac{1}{u} = 0.005 - 0.004902 = 0.000098 \).
06

Finding \( u \)

To find \( u \), take the reciprocal of \( 0.000098 \): \( u = \frac{1}{0.000098} \approx 10204.08 \text{ mm} \).
07

Converting Back to Desired Units

Since the initial context mentioned distances in centimeters, convert \( u \) back to centimeters: \( 10204.08 \text{ mm} = 1020.408 \text{ cm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Focal Length
The focal length is a critical parameter in lens-based optical systems, such as cameras. It is the distance between the lens and its focal point, where light rays converge to form a sharp image. In this exercise, the focal length is
  • 200 mm, a measurement commonly used in describing how strong the lens "pulls" light together.
A lens with a shorter focal length bends light more sharply, bringing it to focus quickly and producing a wider field of view. Conversely, a longer focal length offers a narrower field but is ideal for magnifying distant subjects.
This property directly influences the distance over which the lens can effectively focus an image and plays a crucial role in calculating how far an object must be to be captured sharply by the camera.
Optics
Optics is the branch of physics that deals with the behavior and properties of light. It's fundamental for understanding how lenses like those in cameras work to form images. Inside a lens, light refraction occurs because different surfaces change the speed of light. This change bends light rays, redirecting them to converge at the focal point.
To quantify this, we rely on the lens formula:\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]This formula highlights the relationship between the focal length \( f \), image distance \( v \), and object distance \( u \). Understanding these relationships allows photographers and designers to manipulate light to capture clear images, whether through cameras or other optical devices. Optics goes beyond just lenses, impacting everything from everyday glasses to advanced astronomical telescopes.
Unit Conversion
When working in optics, it's crucial to maintain consistent units. In our example, the focal length is given in millimeters, requiring all other measurements to be in the same unit to effectively use the lens formula. Distances can often be given in a mix of units such as centimeters or meters, and converting is straightforward:
  • To convert centimeters to millimeters, multiply by 10 (since 1 cm equals 10 mm).
For instance, the film's distance from the lens was initially given as 20.4 cm:
  • Converted, this is 204 mm (20.4 cm \( \times \) 10).
Keeping units consistent is like speaking the same language—a necessary step to ensure calculations are accurate and logical.
Distance Measurement
In optical physics, distance measurement between elements like an object and a lens or a lens and its image is pivotal. Here, using the lens formula, we identify the required distance to achieve sharp focus. Calculating the object's distance from the lens involves rearranging the formula:\[ \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \]This step simplifies arraying distances in terms
  • of how far an object must be from the lens (\( u \)) to be imaged correctly.
After substituting given distances, calculations yield \( u \) as approximately 10204.08 mm or 1020.408 cm. Understanding distance measurement allows photographers to creatively decide where to place a subject relative to the lens to achieve the desired effect, leveraging the mathematics of optics.

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Most popular questions from this chapter

An An LCD projector (see Sec. 25.2\()\) has a projection lens with \(f\) -number of 1.8 and a diameter of 46 \(\mathrm{mm}\) . The LCD array measures 3.30 \(\mathrm{cm} \times 3.30 \mathrm{cm}\) and will be projected on a screen 8.00 m from the lens. If the array is \(800 \times 600\) pixels, what will be the dimensions of a single pixel on the screen?

\bullet A microscope with an objective of focal length 8.00 \(\mathrm{mm}\) and an eyepiece of focal length 7.50 \(\mathrm{cm}\) is used to project an image on a screen 2.00 \(\mathrm{m}\) from the eyepiece. Let the image distance of the objective be 18.0 \(\mathrm{cm} .\) (a) What is the lateral magnification of the image? (b) What is the distance between the objective and the eyepiece?

You are designing a projection system for a hall having a screen measuring 4.00 m square. The lens of a 35 \(\mathrm{mm}\) slide projector in the projection booth is 15.0 \(\mathrm{m}\) from this screen. You want to focus the image of 35 \(\mathrm{mm}\) slides (which are 24 \(\mathrm{mm} \times 36 \mathrm{mm}\) ) onto this screen so that you can fill as much of the screen as possible without any part of the image extending beyond the screen. (a) What focal-length lens should you use in the projector? (b) How far from the lens should the slide be placed? (c) What are the dimensions of the slide's image on the screen?

. The focal length of a simple magnifier is 8.00 \(\mathrm{cm} .\) Assume the magnifier to be a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observer's near point, 25.0 \(\mathrm{cm}\) in front of her eye? (b) If the object is 1.00 \(\mathrm{mm}\) high, what is the height of its image formed by the magnifier?

Your digital camera has a lens with a 50 \(\mathrm{mm}\) focal length and a sensor array that measures 4.82 \(\mathrm{mm} \times 3.64 \mathrm{mm}\) . Suppose you're at the zoo, and want to take a picture of a \(4.50-\mathrm{m}-\) tall giraffe. If you want the giraffe to exactly fit the longer dimension of your sensor array, how far away from the animal will you have to stand?

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