/*! 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 35 A camera lens has a focal length... [FREE SOLUTION] | 91Ó°ÊÓ

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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 object should be approximately 1020.41 cm from the lens.

Step by step solution

01

Understand the Problem

We need to find the distance between the lens and the object (subject for the photo). We are given the focal length of the lens and the distance from the lens to the film. We will use the lens formula to find this distance.
02

Apply 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. Here \( f = 200 \; \mathrm{mm} = 20 \; \mathrm{cm} \) and \( d_i = 20.4 \; \mathrm{cm} \). We need to find \( d_o \).
03

Substitute Known Values

Substitute the given values into the lens formula: \( \frac{1}{20} = \frac{1}{d_o} + \frac{1}{20.4} \).
04

Simplify the Equation

To find \( \frac{1}{d_o} \), rearrange the equation: \( \frac{1}{d_o} = \frac{1}{20} - \frac{1}{20.4} \). Calculate each term on the right-hand side.
05

Perform the Calculation

Calculate \( \frac{1}{20} = 0.05 \) and \( \frac{1}{20.4} \approx 0.04902 \). Then, \( \frac{1}{d_o} = 0.05 - 0.04902 = 0.00098 \).
06

Solve for Object Distance

Solve for \( d_o \) by taking the reciprocal of \( 0.00098 \): \( d_o \approx 1020.41 \; \mathrm{cm} \).

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Key Concepts

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

Focal Length
In optics, the focal length of a lens is a crucial concept that helps determine how the lens will focus light. It is defined as the distance from the lens to the point where parallel rays of light converge to a single point, known as the focus. This measurement is crucial because it tells us how strongly the lens converges or diverges light.
  • A shorter focal length indicates a stronger lens, which bends light more and focuses it in a shorter distance.
  • A longer focal length means a weaker lens that bends light less.
In practical terms, the focal length affects image size and field of view. A shorter focal length lens captures a wider area but with less magnification, while a longer focal length lens captures a narrower field with greater detail or magnification. For the camera lens in our exercise, having a focal length of 200 mm (or 20 cm) means it can capture detailed, close-up images from a more considerable distance.
Object Distance Calculation
In optical systems, the object distance is the distance from the lens to the object being imaged. Calculating this distance is essential for properly focusing a camera. This is where the lens formula comes in handy:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Here, - \( f \) is the focal length,- \( d_o \) is the object distance, and- \( d_i \) is the image distance.Given \( f = 20 \; \text{cm} \) and \( d_i = 20.4 \; \text{cm} \), our goal is to find \( d_o \). Start by isolating \( \frac{1}{d_o} \):\[\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}\]By calculating each term, we arrive at \( \frac{1}{20} = 0.05 \) and \( \frac{1}{20.4} \approx 0.04902 \). Thus,\[ \frac{1}{d_o} = 0.05 - 0.04902 = 0.00098 \]To find the object distance \( d_o \), take the reciprocal of 0.00098:\[ d_o \approx 1020.41 \; \text{cm} \].This result tells us where to place the subject relative to the lens to achieve a clear image.
Image Distance
Image distance refers to the distance from the lens to the image produced. In a camera, this is often similar to the distance from the lens to the film or sensor. Understanding image distance is key for focusing a camera or any optical device so that the image is sharp and clear.In the provided example, the image distance \( d_i \) is given as 20.4 cm. This is the distance at which the lens projects a sharp, in-focus image onto the film.
  • A shorter image distance usually means that the image is more zoomed in.
  • Conversely, a longer image distance suggests the image is captured with a wider field of view.
By using the lens formula in the exercise, you can adjust either the object or image distance to maintain a sharp focus, a fundamental concept in photography and optics.

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Most popular questions from this chapter

Sketch the various possible thin lenses that can be obtained by combining two surfaces whose radii of curvature are 4.00 \(\mathrm{cm}\) and 8.00 \(\mathrm{cm}\) in absolute magnitude. Which are converging and which are diverging? Find the focal length of each if the surfaces are made of glass with index of refraction 1.60 .

Resolution of a Microscope. The image formed by a microscope objective with a focal length of 5.00 \(\mathrm{mm}\) is 160 \(\mathrm{mm}\) from its second focal point. The eyepiece has a focal length of 26.0 \(\mathrm{mm}\) . (a) What is the angular magnification of the microscope? (b) The unaided eye can distinguish two points at its near point as separate if they are about 0.10 \(\mathrm{mm}\) apart. What is the minimum separation between two points that can be observed (or resolved) through this microscope?

A spherical, concave, shaving mirror has a radius of curvature of \(32.0 \mathrm{cm} .\) (a) What is the magnification of a person's face when it is 12.0 \(\mathrm{cm}\) to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virual? (c) Draw a principal-ray diagram showing the formation of the image.

A candle 4.85 \(\mathrm{cm}\) tall is 39.2 \(\mathrm{cm}\) to the left of a plane mirror. Where is the image formed by the mirror, and what is the height of this image?

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.

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