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An object is \(30.0 \mathrm{~cm}\) from a spherical mirror, along the mirror's central axis. The mirror produces an inverted image with a lateral magnification of absolute value \(0.500 .\) What is the focal length of the mirror?

Short Answer

Expert verified
The focal length of the mirror is 10.0 cm.

Step by step solution

01

Understanding the problem

We have an object placed 30.0 cm from a spherical mirror, and it produces an inverted image with a lateral magnification of absolute value 0.500. We need to find the focal length of the mirror.
02

Use the magnification formula

The magnification (\(m\)) is given by the formula \(m = -\frac{q}{p}\), where \(q\) is the image distance and \(p\) is the object distance. We know \(m = -0.500\) and \(p = 30.0\, \text{cm}\). Substitute these values to find \(q\).
03

Calculate the image distance

\(m = -0.500\) implies \(-0.500 = -\frac{q}{30.0}\). Solving for \(q\), we get \(q = 15.0\, \text{cm}\).
04

Apply the mirror equation

The mirror equation is \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\), where \(f\) is the focal length. Substitute \(p = 30.0\, \text{cm}\) and \(q = 15.0\, \text{cm}\) into the equation.
05

Calculate the focal length

Substitute \(p = 30.0\, \text{cm}\) and \(q = 15.0\, \text{cm}\) into \(\frac{1}{f} = \frac{1}{30.0} + \frac{1}{15.0}\). Calculate \(\frac{1}{f} = \frac{1}{30.0} + \frac{2}{30.0} = \frac{3}{30.0}\). Simplifying gives \(\frac{1}{f} = \frac{1}{10.0}\). So, \(f = 10.0\, \text{cm}\).

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Key Concepts

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

Understanding Lateral Magnification
Lateral magnification is a key concept when talking about mirrors, particularly spherical mirrors. This concept tells us how much larger or smaller an image is compared to the actual object creating it. In mathematical terms, lateral magnification (\(m\)) can be expressed as the ratio of the image distance (\(q\)) to the object distance (\(p\)). The formula usually used is given by \[ m = -\frac{q}{p} \].
The negative sign in the formula is crucial; it indicates that the image formed is inverted relative to the object. This helps when you know whether your image has flipped or not.
In practical terms:
  • If the absolute value of \(m\) is less than 1, the image is smaller than the object.
  • If the absolute value of \(m\) is greater than 1, the image is larger.
  • If \(m\) is positive, the image is upright.
  • If \(m\) is negative, the image is inverted.
In this problem, we were given a magnification of -0.500, indicating the image is half the size of the object and inverted.
Exploring Focal Length
When dealing with spherical mirrors, the focal length is an essential characteristic. It tells us how strongly the mirror converges or diverges light. The focal length (\(f\)) is the distance from the mirror to the focal point, where parallel rays would converge (or appear to diverge).
To find the focal length, you need to remember:
  • Convex mirrors have positive focal lengths.
  • Concave mirrors have negative focal lengths for real images and positive for virtual images.
Given the equation for finding focal length: \[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \], we use values of object distance (\(p\)) and image distance (\(q\)) to adjust the equation to solve for \(f\).
In this case, with an object distance of 30 cm and image distance of 15 cm, we find the focal point is determined by rearranging the mirror equation.
Using the Mirror Equation
The mirror equation is a powerful tool in optics that relates three main variables: the object distance (\(p\)), the image distance (\(q\)), and the focal length (\(f\)) of a spherical mirror. This equation is crucial for calculating how an image will be formed and its characteristics.
The mirror equation is expressed as:\[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \].
This equation helps us compute the unknown variable when the other two are known.Consider these key steps when applying the mirror equation:
  • Identify the values you have: object distance \(p\), image distance \(q\), or focal length \(f\).
  • Substitute these values into the equation with consistent units.
  • Calculate the missing value by simplifying the expression.
In the example given, the known values were \(p = 30\) cm and \(q = 15\) cm. Using the mirror equation, we calculated \(f\) to be 10 cm, which reveals how this particular mirror bends light to form images.

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Most popular questions from this chapter

A millipede sits \(1.0 \mathrm{~m}\) in front of the nearest part of the surface of a shiny sphere of diameter \(0.70 \mathrm{~m}\). (a) How far from the surface does the millipede's image appear? (b) If the millipede's height is \(2.0 \mathrm{~mm},\) what is the image height? (c) Is the image inverted?

An object is \(20 \mathrm{~cm}\) to the left of a thin diverging lens that has a \(30 \mathrm{~cm}\) focal length. (a) What is the image distance \(i ?\) (b) Draw a ray diagram showing the image position.

The formula \(1 / p+1 / i=1 / f\) is called the Gaussian form of the thin-lens formula. Another form of this formula, the Newtonian form, is obtained by considering the distance \(x\) from the object to the first focal point and the distance \(x^{\prime}\) from the second focal point to the image. Show that \(x x^{\prime}=f^{2}\) is the Newtonian form of the thin-lens formula.

Object \(O\) stands on the central axis of a thin symmetric lens. For this situation, each problem in Table \(34-6\) gives object distance \(p\) (centimeters), the type of lens (C stands for converging and D for diverging), and then the distance (centimeters, without proper sign) between a focal point and the lens. Find (a) the image distance \(i\) and (b) the lateral magnification \(m\) of the object, including signs. Also, determine whether the image is (c) real (R) or virtual (V), (d) inverted (I) from object \(O\) or noninverted (NI), and (e) on the same side of the lens as object \(O\) or on the opposite side.

Someone with a near point \(P_{n}\) of \(25 \mathrm{~cm}\) views a thimble through a simple magnifying lens of focal length \(10 \mathrm{~cm}\) by placing the lens near his eye. What is the angular magnification of the thimble if it is positioned so that its image appears at (a) \(P_{n}\) and (b) infinity?

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