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Chapter 34: Problem 106

A camera with a 90 -mm-focal-length lens is focused on an object 1.30 \(\mathrm{m}\) from the lens. To refocus on an object 6.50 \(\mathrm{m}\) from the lens, by how much must the distance between the lens and the film be changed? To refocus on the more distant object, is the lens moved toward or away from the film?

Short Answer

Expert verified
The lens moves 9.58 mm towards the film to refocus.

Step by step solution

01

Understand the Problem

We need to determine the change in lens-to-film distance when the camera lens refocuses from a near to a far object. The necessary information includes: the focal length (90 mm), the initial object distance (1.30 m), and the new object distance (6.50 m). We will use the lens formula for this purpose.
02

Convert Units for Consistency

Convert the given object distances from meters to millimeters to match the focal length unit: - Initial object distance: 1.30 m = 1300 mm - New object distance: 6.50 m = 6500 mm.
03

Apply the Lens Formula for Initial Object

The lens formula is: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \).- For the initial object, \( f = 90 \text{ mm} \), \( u = 1300 \text{ mm} \).- Rearrange to find the image distance \( v \): \[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{90} - \frac{1}{1300} \].- Calculate \( v \) using the above values to find the initial image distance.
04

Calculate Initial Image Distance

Using the rearranged lens formula: \[ \frac{1}{v} = \frac{1300 - 90}{117000} \ {\Rightarrow} \ v \approx 100.86 \text{ mm} \] So, the image distance \( v \) for the initial object is approximately 100.86 mm.
05

Apply the Lens Formula for New Object Distance

Now, repeat the process for the new object distance:- New object distance, \( u = 6500 \text{ mm} \).- Use the lens formula: \[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{90} - \frac{1}{6500} \].- Calculate the new image distance \( v \).
06

Calculate New Image Distance

Calculate \( v \) using the rearranged formula: \[ \frac{1}{v} = \frac{6500 - 90}{585000} \ {\Rightarrow} \ v \approx 91.28 \text{ mm} \]So, the new image distance \( v \) for the new object is approximately 91.28 mm.
07

Determine the Change in Image Distance

Find the difference between the two image distances:- Change in image distance \( = |100.86 \text{ mm} - 91.28 \text{ mm}| \).- Calculate the change: \( \ = 9.58 \text{ mm} \).
08

Determine Direction of Lens Movement

Since the image distance decreased from 100.86 mm to 91.28 mm, the lens has moved closer to the film to refocus on the more distant object.

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Key Concepts

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

Lens Formula
The lens formula is a fundamental equation in optics that relates the focal length of a lens to the distances of the object and the image it forms. This relationship is given by the formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \) where \( f \) is the focal length, \( v \) is the image distance (distance from the lens to the image), and \( u \) is the object distance (distance from the lens to the object).
This formula helps us determine how lenses focus light to form clear images. It is crucial in applications like photography, where precise focus adjustment is needed. Understanding this formula allows us to predict how changes in one parameter affect the others, helping us adjust lenses to capture clear images at different distances.
Focal Length
The focal length of a lens is a measure of how strongly it converges or diverges light. It is the distance between the lens and the point where parallel rays of light converge to a focus. In the context of the lens formula, the focal length \( f \) is a fixed value determined by the physical characteristics of the lens.
In a camera lens, the focal length can influence the field of view and magnification. A shorter focal length means a wider field of view, capturing more of the scene, while a longer focal length provides higher magnification. For our exercise, the lens had a focal length of 90 mm, which was consistent throughout the problem, influencing how we calculated the image distance for varying object distances.
Image Distance
The image distance, denoted by \( v \) in the lens formula, is the distance from the lens to the image formed. It is one of the variables in determining how an image focuses based on object distance and the lens's focal length.
Calculating the image distance is crucial in focusing a camera. For the initial object (1.3 m), the image distance was found to be approximately 100.86 mm. When refocusing the lens for an object at 6.5 m, the image distance became roughly 91.28 mm. Understanding how the image distance changes with varying object distances helps in adjusting the lens, ensuring the image captured is sharp and clear.
Object Distance
The object distance \( u \), another key element of the lens formula, represents how far the object is from the lens. Changes in the object distance can significantly alter the image distance due to the fixed nature of the focal length.
In the exercise, the object distance initially was 1.30 m (1300 mm) and later changed to 6.50 m (6500 mm). By applying the lens formula, we saw how this shift affected the image distance, prompting adjustments in the lens-to-film setup. A thorough grasp of how object distance impacts the overall optics system enables precise focusing by making the necessary physical adjustments to the lens position.

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Most popular questions from this chapter

The focal length of a simple magnifier is 8.00 \(\mathrm{cm} .\) Assume the magnifier is 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 mm high, what is the height of its image formed by the magnifier?

What Is the Smallest Thing We Can See? The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns \((\mu \mathrm{m})\) is typical near the center of the eye. We shall model the eye as a sphere 2.50 \(\mathrm{cm}\) in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells 5.0\(\mu \mathrm{m}\) in diameter. (a) What is the smallest object you can resolve at a near point of 25 \(\mathrm{cm}\) ? (b) What angle is subtended by this object at the eye? Express your answer in units of minutes \(\left(1^{\circ}=60 \mathrm{min}\right),\) and compare it with the typical experimental value of about 1.0 min. (Note: There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30.0 \(\mathrm{cm}\) to the right of the lens. A diverging lens is now placed 15.0 \(\mathrm{cm}\) to the right of the converging lens, and it is found that the screen must be moved 19.2 \(\mathrm{cm}\) farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

A glass rod with a refractive index of 1.55 is ground and polished at both ends to hemispherical surfaces with radii of 6.00 \(\mathrm{cm} .\) When an object is placed on the axis of the rod, 25.0 \(\mathrm{cm}\) to the left of the left-hand end, the final image is formed 65.0 \(\mathrm{cm}\) to the right of the right-hand end. What is the length of the rod measured between the vertices of the two hemispherical surfaces?

Focal Length of a Zoom Lens. Figure P34.113 shows a simple version of a zoom lens. The converging lens has focal length \(f_{1},\) and the diverging lens has focal length \(f_{2}=-\left|f_{2}\right|\) The two lenses are separated by a variable distance \(d\) that is always less than \(f_{1} .\) Also, the magnitude of the focal length of the diverging lens satisfies the inequality \(\left|f_{2}\right|>\left(f_{1}-d\right) .\) To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius \(r_{0}\) entering the converging lens. (a) Show that the radius of the ray bundle decreases to \(r_{0}^{\prime}=r_{0}\left(f_{1}-d\right) / f_{1}\) at the point that it enters the diverging lens. (b) Show that the final image \(I^{\prime}\) is formed a distance \(s_{2}^{\prime}=\left|f_{2}\right|\left(f_{1}-d\right) /\left(\left|f_{2}\right|-f_{1}+d\right)\) to the right of the diverging lens. (c) If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left of the diverging lens, they will eventually expand to the original radius \(r_{0}\) at some point \(Q .\) The distance from the final image \(I^{\prime}\) to the point \(Q\) is the effective focal length \(f\) of the lens combination; if the combination were replaced by a single lens of focal length \(f\) placed at \(Q,\) parallel rays would still be brought to a focus at \(I^{\prime}\) . Show that the effective focal length is given by \(f=f_{1}\left|f_{2}\right| /\left(\left|f_{2}\right|-f_{1}+d\right) .\) (d) If \(f_{1}=12.0 \mathrm{cm}\) \(f_{2}=-18.0 \mathrm{cm},\) and the separation \(d\) is adjustable between 0 and \(4.0 \mathrm{cm},\) find the maximum and minimum focal lengths of the combination. What value of \(d\) gives \(f=30.0 \mathrm{cm} ?\)
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