Question

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

The focal length of the lens is 10.0 cm.

Step-by-step solution

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.

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, \( d_i = 30.0 \text{ cm} \).

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 \( d_i' = 30.0 - 4.0 - 4.0 = 22.0 \text{ cm} \).

Calculate Object Distance Change

The object distance is increased by the movement of the lens. Therefore, the new object distance \( d_o' = d_o + 4.0 \).

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.

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, \( d_o = 15.0 \text{ cm} \).

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, \( f = 10.0 \text{ cm} \).

Key concepts

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.

This equation helps find the relationship between these distances and the lens's focal length. It ties together how the position of an object and its image shifts when the lens and screen are adjusted. Understanding this formula is crucial, as it is widely used in optical calculations to solve real-world problems.

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:

• 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.

Initially, the lens movement requirement helped establish that the object distance changes. Through our calculations, we determined an object distance of 15.0 cm. This value allows us to apply the thin lens formula to find the focal length accurately.

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.

To refocus the image on a screen, this adjustment in distance was necessary. By understanding the changes in image distance, the lens’s optical properties are more evident, aiding in deriving the correct focal length from the problem.