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Photography. A 35 -mm camera has a standard lens with focal length 50 \(\mathrm{mm}\) and can focus on objects between 45 \(\mathrm{cm}\) and infinity. (a) Is the lens for such a camera a concave or a convex lens? (b) The camera is focused by rotating the lens, which moves it on the camera body and changes its distance from the film. In what range of distances between the lens and the film plane must the lens move to focus properly over the 45 \(\mathrm{cm}\) to infinity range?

Short Answer

Expert verified
(a) The lens is a convex lens. (b) The lens moves between 50 mm and 56.25 mm from the film plane.

Step by step solution

01

Understand Lens Type

In cameras, the lens helps converge light to form a clear image. A 50 mm focal length in a 35mm camera is considered a standard lens. For cameras, the lenses need to converge light, pointing to a convex lens, as concave lenses diverge light.
02

Use Lens Formula

The lens formula is given by \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \), where \( f \) is the focal length, \( u \) is the object distance, and \( v \) is the image distance. In our case, \( f = 50 \) mm.
03

Calculate Image Distance for Closest Focusing Distance

Set the object distance \( u = 450 \) mm (as 45 cm = 450 mm). Rearranging the lens formula, \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \), substitute \( f = 50 \) mm and \( u = 450 \) mm, then solve for \( v \).\[ \frac{1}{v} = \frac{1}{50} - \frac{1}{450} = \frac{450 - 50}{22500} = \frac{400}{22500} \] \( v = \frac{22500}{400} = 56.25 \) mm.
04

Calculate Image Distance for Focusing at Infinity

When the camera is focused on infinity, \( u = \infty \). Thus, \( \frac{1}{u} = 0 \), reducing the lens formula to \( \frac{1}{v} = \frac{1}{f} \). Hence \( v = f = 50 \) mm.
05

Determine Range of Lens Movement

Comparing the image distances calculated for closest (56.25 mm) and infinity (50 mm) focus, the lens must be capable of moving between these two positions. Therefore, the lens needs to move between 50 mm and 56.25 mm from the film plane.

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Key Concepts

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

Convex Lens
In the world of photography, a convex lens plays a crucial role. A convex lens is a type of lens that is thicker at the center than at the edges. This shape causes light rays to converge at a point. For cameras, this is essential for forming sharp images on the film or sensor.
Convex lenses are the go-to choice for most photographic applications because they help gather and focus light from objects, creating clear images. Without this lens type, photographs would appear blurry.
In general, a convex lens is used in various optical devices like telescopes and glasses due to its light converging property.
  • Convex lenses converge light rays to meet at a certain point.
  • Essential for creating sharp camera images.
  • Used in various optical devices.
Lens Formula
The lens formula is essential in photography and optics. It helps to relate the object distance, image distance, and focal length. This formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
Where:
- \( f \) is the focal length,- \( u \) is the object distance,- \( v \) is the image distance.
Understanding this formula is important because it aids photographers in knowing how to adjust their camera settings to focus on different object distances. By rearranging this formula, photographers can calculate how the lens should be adjusted to focus correctly.
Whether you're aiming to capture distant landscapes or close-up portraits, the lens formula serves as the foundation for understanding how the lens behaves in different focusing situations.
Image Distance
The term 'image distance' refers to the distance between the lens and the image formed on the film or sensor. It's important because it determines where the camera lens should be positioned to achieve a clear image.
When using the lens formula, if you want to find the image distance \( v \), you rearrange the equation to get \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \).
For example, if your camera is focusing on an object 45 cm away, you plug the numbers into the formula to find the image distance. The calculation reveals that the lens must be at a specific position from the film to ensure clarity.
This understanding allows photographers to adjust the lens movement between the closest focus and infinity focus, as seen in cameras with zoom or focus adjustments.
  • Image distance helps determine lens positioning.
  • Critical for achieving clear, focused images.
  • Connects to both focal length and object distance in lens formula.
Focal Length
Focal length is a key characteristic of a lens and defines the field of view of a camera. It's the distance from the center of the lens to the focal point, where light rays converge to form a sharp image.
For example, a 50 mm lens is often standard in many cameras as it offers a natural perspective similar to human vision. A shorter focal length means a wider field of view, while a longer one results in a narrow and more zoomed-in view.
Photographers need to understand focal length to select the right lens for their subject matter. It influences both the composition of the photograph and the detail captured.
The focal length also plays a crucial role in the lens formula, influencing both the object and image distances. Understanding how focal length affects image capture is essential for anyone looking to effectively use a camera.
  • Focal length impacts field of view and perspective.
  • Important for lens selection based on subject.
  • Integral part of the lens formula.

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Most popular questions from this chapter

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_{1}=+12.0 \mathrm{cm}\) and \(R_{2}=+28.0 \mathrm{cm} .\) The index of refraction is 1.60 . (a) Compute the position and size of the image of an object in the form of an arrow 5.00 \(\mathrm{mm}\) tall, perpendicular to the lens axis, 45.0 \(\mathrm{cm}\) to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 \(\mathrm{m}\) to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens 45.0 \(\mathrm{cm}\) to the right of the first.

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?

The Cornea As a Simple Lens. The cornea behaves as a thin lens of focal length approximately \(1.8 \mathrm{cm},\) although this varies a bit. The material of which it is made has an index of refraction of \(1.38,\) and its front surface is convex, with a radius of curvature of 5.0 \(\mathrm{mm} .\) (a) If this focal length is in air, what is the radius of curvature of the back side of the cornea? (b)The closest distance at which a typical person can focus on an object (called the near point) is about \(25 \mathrm{cm},\) although this varies considerably with age. Where would the cornea focus the image of an 8.0 -mm- tall object at the near point? (c) What is the height of the image in part (b)? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because, on one side, the cornea has a fluid with a refractive index different from that of air.)

If you run away from a plane mirror at \(3.60 \mathrm{m} / \mathrm{s},\) at what speed does your image move away from you?

A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of 18.0 \(\mathrm{cm} .\) Reflection from the surface of the shell forms an image of the \(1.5-\) cm-tall coin that is 6.00 \(\mathrm{cm}\) behind the glass shell. Where is the coin located? Determine the size, orientation, and nature (real or virtual) of the image.

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