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Chapter 19: Problem 66

A concave mirror produces an image \(n\) times the size of an object. If the focal length of the mirror is ' \(f^{\prime}\) and image formed is real, then the distance of the object from the mirror is : (a) \((n-1) f\) (b) \(\frac{(n-1)}{n} f\) (c) \(\frac{(n+1)}{n} f\) (d) \((n+1) f\)

Short Answer

Expert verified
The object is at a distance \(\frac{(n+1)}{n} f\) from the mirror.

Step by step solution

01

Understand the Magnification Formula for Mirrors

The magnification (\(m\)) produced by a mirror is defined as the ratio of the height of the image (\(h'\)) to the height of the object (\(h\)), which can also be expressed in terms of image distance (\(v\)) and object distance (\(u\)): \(m = -\frac{v}{u}\). In this scenario, the magnification is \(n\), which means \(-\frac{v}{u} = n\).
02

Resolve Object and Image Distance

From \(-\frac{v}{u} = n\), we can express \(v\) in terms of \(u\): \(v = -nu\). Since the image is real for a concave mirror, both \(v\) and \(u\) are negative; hence \(v = nu\).
03

Apply Mirror Formula

The mirror formula is \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\). Substituting \(v = nu\) into this, we have \(\frac{1}{f} = \frac{1}{u} + \frac{1}{nu}\).
04

Simplify the Equation

Factor \(\frac{1}{u}\) from the right-hand side: \(\frac{1}{f} = \frac{1}{u} \left( 1 + \frac{1}{n} \right)\). Simplify to get \(\frac{1}{u} = \frac{n}{f(n+1)}\).
05

Determine Object Distance from Focal Length

Isolate \(u\) in the equation \(u = \frac{f(n+1)}{n}\). This indicates the distance of the object from the mirror in terms of given focal length and magnification factor \(n\).
06

Identify the Correct Option

Comparing \(u = \frac{f(n+1)}{n}\) with the options provided in the exercise, we find that option (c) \(\frac{(n+1)}{n} f\) matches our derived formula.

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Key Concepts

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

Magnification
When dealing with concave mirrors, magnification is a crucial concept. It tells us how much larger or smaller an image is compared to the actual object. Magnification (\(m\)) is defined as the ratio of the image height (\(h'\)) to the object height (\(h\)).
However, in terms of concave mirrors, it's often more useful to express magnification through distances: \(m = -\frac{v}{u}\). Here, \(v\) represents the image distance, and \(u\) is the object distance from the mirror.
  • The negative sign in the formula indicates the nature of the image: a real image will have a negative image distance, following the sign convention for mirrors.
  • If the magnification is positive, the image is virtual and erect. If negative, the image is real and inverted.
Understanding magnification helps us determine how the image will appear relative to the object's size.
Mirror Formula
The mirror formula is a fundamental equation that applies to spherical mirrors, such as concave mirrors. It connects three essential variables: the focal length, image distance, and object distance. This equation is given by:
\(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\)
  • \(f\) is the focal length of the mirror, which is half the radius of curvature.
  • \(u\) is the distance from the object to the mirror.
  • \(v\) is the distance from the mirror to the image.
Understanding this formula allows you to find any of the three variables if the other two are known. It's particularly useful in situations involving complex scenarios, like determining one when the mirror is changing positions.
Object Distance
The concept of object distance (\(u\)) plays a vital role in understanding mirror reflections. It is the distance from the object to the pole of the mirror and, in concave mirrors, is treated as a negative value.
  • The sign convention dictates that all distances measured in the direction of the incoming light are negative for concave mirrors.
  • In problems, object distance helps establish the position where the object needs to be placed to obtain a desired image type, such as real vs. virtual, magnified, or diminished.
In this exercise, by rearranging the mirror formula and considering the magnification factor, we found that the object distance \(u\) is \(\frac{f(n+1)}{n}\). Fully understanding how object distance relates to focal length and magnification assists in predicting and explaining the behavior of images produced by concave mirrors.
Real Image Formation
Real image formation is one of the defining characteristics of concave mirrors, where the image is formed by actually converging rays. Several key points help describe real images:
  • Real images are inverted compared to the original object.
  • They can be projected onto a screen, unlike virtual images which cannot.

In this context, calculations showed a real image formation when using the given mirror formula combined with object distance variables. The exercise demonstrated that when the magnification was given as a positive value \(n\), and using the real nature of the image, both \(v\) and \(u\) turned out to be negative, indicating a real image. This concept is crucial as it not only helps in traditional problem-solving but aids in a deeper understanding of how light interacts with concave mirrors.

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Most popular questions from this chapter

Two plane mirrors are inclined at an angle such that a ray incident on a mirror undergoes a total deviation of \(240^{\circ}\) after two reflections. The angle between mirrors. Also discuss the formation of image : (a) \(60^{\circ}, 5\) (b) \(5^{\circ}, 60\) (c) \(45^{\circ}, 5\) (d) \(30^{\circ}, 6\)

Images in spherical mirrors suffer from several defects. Some of which is/are: (a) diffraction effect (b) the magnification varies with the distance of the object from mirror (c) a point source will not produce a point image (d) all of the above

A narrow beam of light after reflection by a plane mirror falls on a scale \(100 \mathrm{~cm}\) from the mirror. When the mirror is rotated a little the spot moves through \(2 \mathrm{~cm}\), the angle, through which the mirror is rotated is : (a) \(0.02 \mathrm{rad}\) (b) \(0.01 \mathrm{rad}\) (c) \(200 \mathrm{rad}\) (d) \(\frac{0.01}{180} \pi \mathrm{rad}\)

A convex driving mirror of focal length \(20 \mathrm{~cm}\), is fitted in a motor car. If the second car \(2 \mathrm{~m}\) broad and \(1.6 \mathrm{~m}\) high is \(6 \mathrm{~m}\) away from first car and overtakes the first car at a relative speed of \(15 \mathrm{~m} / \mathrm{s}\), then how fast will the image be moving? (a) \(0.016 \mathrm{~m} / \mathrm{s}\) (b) \(0.0257 \mathrm{~m} / \mathrm{s}\) (c) \(0.162 \mathrm{~m} / \mathrm{s}\) (d) \(0.0073 \mathrm{~m} / \mathrm{s}\)

A ray is incident on a plane surface. If \(\hat{i}+\hat{j}-\hat{k}\) represents a vector along the direction of incident ray. \(\hat{i}+\hat{j}\) is a vector along normal on incident point in the plane of incident and reflected ray. Then vector along the direction of reflected ray is: (a) \(-\frac{1}{\sqrt{19}}(-3 \hat{\mathbf{i}}+3 \hat{j}+\mathbf{k})\) (b) \(\frac{1}{\sqrt{19}}(3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\cdot \hat{\mathrm{k}})\) (c) \(-\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\) (d) \(\hat{\mathbf{k}}\)
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