Chapter 34: Problem 85
An object to the left of a lens is imaged by the lens on a screen 30.0 cm to the right of the lens. When the lens is moved 4.00 cm to the right, the screen must be moved 4.00 cm to the left to refocus the image. Determine the focal length of the lens.
Short Answer
Expert verified
The focal length of the lens is 10.0 cm.
Step by step solution
01
Understand the Problem
We have a lens that initially images an object onto a screen 30.0 cm away. When the lens is moved 4.0 cm right, the screen is moved 4.0 cm left to refocus. We are tasked with finding the focal length of the lens.
02
Apply Thin Lens Formula
Use the thin lens formula, \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. Initially, \( d_i = 30.0 \text{ cm} \).
03
Update Conditions for Lens Movements
When the lens is moved 4.0 cm to the right, to refocus, the screen must be moved 4.0 cm left, making the new image distance \( d_i' = 30.0 - 4.0 - 4.0 = 22.0 \text{ cm} \).
04
Calculate Object Distance Change
The object distance is increased by the movement of the lens. Therefore, the new object distance \( d_o' = d_o + 4.0 \).
05
Relate Initial and New Situations
From the lens equation for both setups, we have:1. \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{30} \)2. \( \frac{1}{f} = \frac{1}{d_o + 4} + \frac{1}{22} \)Since the focal length \( f \) is constant, equate the two scenarios.
06
Solve the System of Equations
From equations in Step 5, solve for \( d_o \) first:\( \frac{1}{d_o} + \frac{1}{30} = \frac{1}{d_o + 4} + \frac{1}{22} \).Simplify and solve:\( \frac{1}{d_o} - \frac{1}{d_o+4} = \frac{1}{22} - \frac{1}{30} \)\( \frac{4}{d_o(d_o+4)} = \frac{8}{660} \)\( d_o^2 + 4d_o = 330 \).By solving this quadratic equation, \( d_o = 15.0 \text{ cm} \).
07
Calculate the Focal Length
Now, use the initial conditions to find the focal length:\( \frac{1}{f} = \frac{1}{15.0} + \frac{1}{30.0} \).Thus, \( \frac{1}{f} = \frac{2 + 1}{30} = \frac{3}{30} \).Therefore, \( f = 10.0 \text{ cm} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Thin Lens Formula
The thin lens formula is essential in understanding how lenses form images. It is given by the equation:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where:
- \( f \) is the focal length of the lens.
- \( d_o \) is the object distance, i.e., the distance from the object to the lens.
- \( d_i \) is the image distance, i.e., the distance from the lens to the image formed.
Focal Length
The focal length \( f \) of a lens is a measure of how strongly the lens converges or diverges light. It is the distance from the lens to the point where it focuses parallel rays of light.
The focal length determines the power of the lens:
In the given problem, we found the focal length to be 10.0 cm. This implies that the lens has a moderate converging power, capable of focusing light to form an image efficiently.
The focal length determines the power of the lens:
- A smaller focal length means the lens has a higher converging power, gathering light more quickly.
- A larger focal length indicates a lens with lower converging power.
In the given problem, we found the focal length to be 10.0 cm. This implies that the lens has a moderate converging power, capable of focusing light to form an image efficiently.
Object Distance
Object distance \( d_o \) refers to the distance between the object and the lens. It's a critical parameter when it comes to forming the image:
- As the object moves closer or further from the lens, the image position and characteristics change.
- In the original problem, calculations start with an unknown object distance, adjusted by the movement of the lens.
Image Distance
Image distance \( d_i \) is the distance between the lens and the image it forms. In optical systems, this value is integral in determining where the image will appear when the object and lens positions are set:
- Adjusting the image distance can bring an image into sharper focus, as seen in the exercise.
- The initial image distance is 30.0 cm, which changes to 22.0 cm when the lens is shifted.