Problem 66
Question
On the axis of a spherical mirror of focal length \(f\), a short linear object of length \(L\) lies on the axis at a distance \(u\) from the mirror. Its image has an axial length \(L^{\prime}\) equal to: (a) \(L\left[\frac{f}{(u-f)}\right]^{1 / 2}\) (b) \(L\left[\frac{(u+f)}{f}\right]^{1 / 2}\) (c) \(L\left[\frac{(u-f)}{f}\right]^{2}\) (d) \(L\left[\frac{f}{(u-f)}\right]^{2}\)
Step-by-Step Solution
Verified Answer
The answer is option (d) \(L\left(\frac{f}{u-f}\right)^2\).
1Step 1: Understand the Mirror Formula
To solve the problem, we start by using the mirror formula which relates object distance \(u\), image distance \(v\), and focal length \(f\). The formula is: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] This formula will help us find the image distance \(v\).
2Step 2: Rearrange the Mirror Formula
Rearrange the mirror formula to find \(v\): \[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \] Solving for \(v\), we get: \[ v = \frac{uf}{u - f} \] This equation gives us the distance from the mirror to the image.
3Step 3: Calculate Magnification
The linear magnification \(m\) of the mirror is given by the ratio of the image distance \(v\) to the object distance \(u\): \[ m = \frac{v}{u} = \frac{uf}{u(u - f)} = \frac{f}{u - f} \] Thus, magnification is expressed as \( \frac{f}{u-f} \).
4Step 4: Determine the Image Axial Length
Using magnification, the axial length of the image \( L' \) is given by: \[ L' = m^2 \cdot L = \left( \frac{f}{u-f} \right)^2 \cdot L \] Therefore, \(L' = L\left(\frac{f}{u-f}\right)^2\).
5Step 5: Identify the Correct Answer from Options
Comparing the expression for \(L'\) to the given options, we see that option (d) \(L\left(\frac{f}{u-f}\right)^2\) matches our calculation.
Key Concepts
Mirror FormulaLinear MagnificationFocal Length
Mirror Formula
In the study of spherical mirrors, a fundamental concept is the mirror formula. This equation establishes a relationship between three crucial variables: object distance (\( u \)), image distance (\( v \)), and focal length (\( f \)). The formula is written as:
\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]This equation is essential in mirror-related problems as it allows us to calculate the unknown distances if at least two values (\( u, v, f \)) are known.
Let's break it down:
\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]This equation is essential in mirror-related problems as it allows us to calculate the unknown distances if at least two values (\( u, v, f \)) are known.
Let's break it down:
- Focal Length (\( f \)): Represents the distance from the mirror's focal point to its surface.
- Object Distance (\( u \)): The distance from the object to the mirror.
- Image Distance (\( v \)): The distance from the mirror to the image.
Linear Magnification
Linear magnification is another key concept in the context of spherical mirrors. It describes how much larger or smaller the mirror makes the image appear relative to the object's actual size. The formula for linear magnification (\( m \)) is:
\[m = \frac{v}{u} \]Substituting the image distance (\( v \)) derived from the mirror formula, it can be expressed as:
\[m = \frac{f}{u-f} \]Here’s what each component means:
\[m = \frac{v}{u} \]Substituting the image distance (\( v \)) derived from the mirror formula, it can be expressed as:
\[m = \frac{f}{u-f} \]Here’s what each component means:
- Image Distance (\( v \)): How far the image is from the mirror.
- Object Distance (\( u \)): Distance between the object and the mirror.
- A magnification greater than 1 means the image is larger than the object.
- Less than 1 means the image is smaller.
- A negative magnification indicates an inverted image.
Focal Length
Focal length is a significant measurement in the analysis of mirrors and lenses. It essentially defines how strongly the mirror converges or diverges light rays:
- Concave Mirror: Positive focal length because it converges light.
- Convex Mirror: Negative focal length as it diverges light.
- Determines how the image is formed relative to the mirror.
- Helps in calculating both image and object distances using the mirror formula.
- Directly influences the magnification and type (real or virtual) of the image.
Other exercises in this chapter
Problem 64
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