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The lens of a camera has a focal length of \(6 \mathrm{~cm}\) and the camera is focused for very distant objects. Through what distance must the lens be moved, and in what direction, when the focus is readjusted for an object \(2 \mathrm{~m}\) from the lens?

Short Answer

Expert verified
The lens should be moved approximately \(0.186 \mathrm{~cm}\) away from the camera to focus on an object \(2 \mathrm{~m}\) away.

Step by step solution

01

Find the image distance for very distant objects

When the camera is focused for very distant objects, the object distance (\(d_o\)) approaches infinity. In this situation, the lens formula becomes: \(\frac{1}{f} = \frac{1}{\infty} + \frac{1}{d_i}\) Since \(f = 6 \mathrm{~cm}\), we can plug this value into the equation and solve for \(d_i\): \(\frac{1}{6} = 0 + \frac{1}{d_i}\)
02

Calculate the initial image distance

Now, we can solve for \(d_i\) to get the initial image distance: \(\frac{1}{d_i} = \frac{1}{6}\) Multiply both sides by 6: \(d_i = 6 \mathrm{~cm}\) So, the initial image distance is 6 cm.
03

Find the image distance when the object is 2 meters from the lens

When the object is 2 meters (200 cm) away, we can use the lens formula to find the new image distance. Plug the focal length and new object distance into the equation: \(\frac{1}{6} = \frac{1}{200} + \frac{1}{d_i}\)
04

Calculate the new image distance

Now, we can solve for the new image distance: \(\frac{1}{d_i} = \frac{1}{6} - \frac{1}{200}\) Find a common denominator and subtract the fractions: \(\frac{1}{d_i} = \frac{194}{1200}\) Now multiply both sides by 1200: \(d_i = \frac{1200}{194}\) After simplifying, we get: \(d_i \approx 6.186 \mathrm{~cm}\) So, the new image distance is approximately 6.186 cm.
05

Calculate the distance the lens must be moved

Now that we have both image distances, we can find the difference between them to determine how far the lens must be moved: Distance to move = New image distance - Initial image distance Distance to move = \(6.186 - 6\) Distance to move ≈ \(0.186 \mathrm{~cm}\) Since the new image distance is greater than the initial image distance, the lens must be moved away from the camera. Therefore, the lens should be moved approximately 0.186 cm away from the camera to focus on an object 2 meters away.

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Key Concepts

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

Understanding Focal Length
In the world of lens optics, focal length plays a pivotal role in determining how light is focused by the lens. The focal length is the distance between the lens and the image sensor when the lens is focused at infinity. It is typically measured in centimeters or millimeters. A shorter focal length indicates a wider field of view, which captures more of the scene.

In the context of our camera exercise, the focal length is given as 6 cm. This means when the lens is set to infinity, the light rays converge exactly 6 cm behind the lens. This is crucial as it provides a base measurement for adjusting focus based on object distance. Understanding focal length helps you grasp how lenses are able to capture different perspectives and depths.
Explaining Image Distance
Image distance is another essential concept in lens optics. It refers to the distance from the lens to the point where the image forms. When using a camera, this distance changes based on where the object you are focusing on is located. For very distant objects, the image distance can generally be approximated to be equal to the focal length because the object distance is very large.

In our problem, when focusing on an object 2 meters away (or 200 cm), we calculated that the image distance became approximately 6.186 cm. This demonstrates how a closer object requires a slight adjustment in the lens position to maintain focus.
The Lens Formula
The lens formula is a handy mathematical equation for understanding how lenses focus light. The formula is \((\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i})\), where \(f\)\ is the focal length, \(d_o\)\ is the object distance, and \(d_i\)\ is the image distance.

This equation helps you calculate how adjustments in object distance affect the image distance and vice versa. It is especially useful in scenarios like our exercise, where you adjust focus for different object distances. By arranging and solving this formula, you can predict how far to move the lens to achieve the right focus.
Camera Focus Adjustment
Adjusting camera focus involves moving the lens to achieve clear images of objects at different distances. When an object moves closer to the camera, like in our exercise, the lens needs to move further from the image sensor. This ensures the image remains sharp.

In practical terms, this means when you focus on an object 2 meters away from the camera with a 6 cm focal length lens, you move the lens away by approximately 0.186 cm. Understanding the mechanics of focus adjustment allows photographers to precisely capture their desired subject without blurriness.

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Most popular questions from this chapter

A telescope is sighted on the image of a scale formed by reflection in a mirror. Both the telescope objective and the scale are \(1 \mathrm{~m}\) from the mirror (see fig) . What is the lateral magnification of the image of the scale formed by the objective, if the objective has a focal length of \(50 \mathrm{~cm}\) ? What should the angular magnification of the ocular be if the scale is to be read as easily through the telescope as it would be by the unaided eye with the scale \(25 \mathrm{~cm}\) away?

A compound microscope has an objective of \(3 \mathrm{~mm}\) focal length. The objective forms an (intermediary) image \(160 \mathrm{~mm}\) beyond the second focal plane of the objective. If the eyepiece alone gives \(20 \mathrm{x}\) magnification (for the eye focused at infinity), what is the total magnification provided by the microscope?

A compound microscope of overall length \(30 \mathrm{~cm}\) consists of two lenses of focal lengths \(1 \mathrm{~cm}\) and \(5 \mathrm{~cm}\). What is the magnifying power of this microscope for a person whose least distance for distinct vision is \(25 \mathrm{~cm} ?\) How far from the objective is the object viewed?

The objective lens of a telescope is \(20 \mathrm{~mm}\) in diameter. Its focal length is \(250 \mathrm{~mm}\), and the eyepiece of the telescope is \(2 \mathrm{~mm}\) in diameter. (a) What is the normal magnification of the telescope? (b) What focal length ocular should be used? (c) Find the position of the exit pupil. (d) What would be the diameter of the exit pupil if an ocular were used which gave a magnification \(50 \%\) in excess of normal? (e) What would be the diameter of the exit pupil if the magnification were \(50 \%\) of normal? Assume all lenses to be thin.

For light of \(620 \mathrm{~nm}\) wavelength, determine (a) the linear wave number and (b) the band width, in units of wavelength, for a Fourier spectrometer which gives a resolution of \(\mathrm{R}=0.4 \mathrm{~cm}^{-1}\).

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