/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 A macroscopic (or macro) lens fo... [FREE SOLUTION] | 91Ó°ÊÓ

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A macroscopic (or macro) lens for a camera is usually a converging lens of normal focal length built into a lens barrel that can be adjusted to provide the additional lens-tofilm distance needed when focusing at very close range. Suppose that a macro lens \((f=\) \(50.0 \mathrm{~mm}\) ) has a maximum lens-to- film distance of \(275 \mathrm{~mm}\). How close can the object be located in front of the lens?

Short Answer

Expert verified
The object can be about 61.1 mm from the lens.

Step by step solution

01

Define the Lens Formula

The lens formula relates the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\). It is given by the equation:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]This formula will be used to find the object distance.
02

Set Values and Solve for \(d_o\)

Insert the known values into the lens formula: \(f = 50.0 \text{ mm}\) and \(d_i = 275 \text{ mm}\). The equation becomes:\[ \frac{1}{50.0} = \frac{1}{d_o} + \frac{1}{275} \]
03

Rearrange the Equation

Isolate \(\frac{1}{d_o}\) on one side by subtracting \(\frac{1}{275}\) from both sides:\[ \frac{1}{d_o} = \frac{1}{50.0} - \frac{1}{275} \]
04

Calculate \(\frac{1}{d_o}\)

Calculate \(\frac{1}{d_o}\) by finding a common denominator and solving the right-hand side:\[ \frac{1}{d_o} = \frac{275 - 50}{50 \times 275} = \frac{225}{13750} \]
05

Determine Object Distance \(d_o\)

To find \(d_o\), take the reciprocal of \(\frac{1}{d_o}\):\[ d_o = \frac{13750}{225} \approx 61.1 \text{ mm} \]Thus, the object can be approximately 61.1 mm away from the lens.

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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 fundamental property of any lens and describes how strongly the lens converges or diverges light. It is represented by the symbol \( f \) and is measured in millimeters (mm) in the context of camera lenses. A shorter focal length indicates a stronger converging lens, which can focus light from closer objects, whereas a longer focal length is often associated with lenses that capture scenes from a distance. Understanding focal length is crucial for photographers, especially when using macro lenses designed for close-up photography. In our exercise, we have a focal length of 50.0 mm, which tells us how this lens will behave in terms of focusing ability.
Converging Lens
A converging lens, also known as a convex lens, is one that brings parallel rays of light to a focus at a point known as the focal point. These lenses are thicker in the middle and thinner at the edges. Converging lenses play a vital role in many optical devices, such as magnifying glasses and cameras.

With macro photography, converging lenses allow for clear, sharp images even at short distances by bending light inwards. Macro lenses typically have smaller focal lengths and are built to focus at shorter object distances, dynamically adjusting to make sharp images from subjects at close range. In the provided exercise, our lens is described as having a normal 50.0 mm focal length, which is characteristic of a converging lens designed for macro photography.
Object Distance
Object distance, denoted as \( d_o \), is the distance between the lens and the object being photographed. It is a key variable in the lens formula that allows photographers to calculate how far or near an object is from the lens. Understanding object distance is essential for focusing and framing the image correctly.

In the exercise, the object distance was calculated using the lens formula:
  • Formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
  • Inserting the known variables and rearranging the equation led to the calculation of \( d_o \) as approximately 61.1 mm
This tells us how close the object can be positioned relative to the lens, which in macro photography, is essential for capturing details.
Image Distance
Image distance, represented as \( d_i \), is the distance from the lens to the image sensor where the focused image is formed. This distance is critical in determining the clarity and focus of the image captured by the camera. When using a fixed setup, image distance needs to be adjusted to maintain focus on subjects at various distances.

For the macro lens in this particular exercise, the maximum image distance is given as 275 mm. By keeping the image distance fixed and using the lens formula, one can compute how close the object can be for it to remain in focus. This is a practical application in photography, especially in macro and close-up shooting scenarios.

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Most popular questions from this chapter

A farsighted person can read printing as close as \(25.0 \mathrm{~cm}\) when she wears contacts that have a focal length of \(45.4 \mathrm{~cm}\). One day, however, she forgets her contacts and uses a magnifying glass, as in Figure \(26-40 b\). It has a maximum angular magnification of 7.50 for a young person with a normal near point of \(25.0 \mathrm{~cm}\). What is the maximum angular magnification that the magnifying glass can provide for her?

Amber \((n=1.546)\) is a transparent brown-yellow fossil resin. An insect, trapped and preserved within the amber, appears to be \(2.5 \mathrm{~cm}\) beneath the surface when viewed directly from above. How far below the surface is the insect actually located?

A ray of sunlight is passing from diamond into crown glass; the angle of incidence is \(35.00^{\circ} .\) The indices of refraction for the blue and red components of the ray are: blue \(\left(n_{\text {diamond }}=2.444, n_{\text {crown glass }}=1.531\right),\) and red \(\left(n_{\text {diamond }}=2.410, n_{\text {crown glass }}=1.520\right)\) Determine the angle between the refracted blue and red rays in the crown glass.

A silver medallion is sealed within a transparent block of plastic. An observer in air, viewing the medallion from directly above, sees the medallion at an apparent depth of \(1.6 \mathrm{~cm}\) beneath the top surface of the block. How far below the top surface would the medallion appear if the observer (not wearing goggles) and the block were under water?

A scuba diver, submerged under water, looks up and sees sunlight at an angle of \(28.0^{\circ}\) from the vertical. At what angle, measured from the vertical, does this sunlight strike the surface of the water?

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