Question

A telescope has been properly focused on the Sun. You want to observe the Sun visually, but to protect your sight you don't want to look through the eyepiece; rather, you want to project an image of the Sun on a screen \(1.5 \mathrm{~m}\) behind (the original position of) the eyepiece, and observe that. If the focal length of the eyepiece is \(8.0 \mathrm{~cm},\) how must you move the eyepiece?

Short answer

In order to project the Sun's image onto a screen 1.5 meters behind the original position of the eyepiece, the eyepiece must be moved 150 cm away from the initial image position (toward the screen). The object distance should approach infinity, as it represents the distance between the eyepiece and the Sun, which is considered to be at an infinite distance.

Step-by-step solution

Write down the lens formula

The lens formula is given by: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] where \(f\) is the focal length of the lens (in this case, the eyepiece), \(d_o\) is the object distance (distance between the lens and the object), and \(d_i\) is the image distance (distance between the lens and the image formed). Our goal is to find how much we need to move the eyepiece, so we are looking for the change in image distance, \(\Delta d_i\).

Find the initial image distance

Since the telescope is properly focused on the Sun, the image distance, \(d_{i1}\), will be equal to the focal length of the eyepiece, which is 8 cm. So, \(d_{i1} = 8 \mathrm{~cm}\).

Find the final image distance

We want to project the image on a screen 1.5 meters (150 cm) behind the original position of the eyepiece. Therefore, the final image distance, \(d_{i2}\), will be the sum of the initial image distance and the screen distance, i.e., \(d_{i2} = d_{i1} + 150 \mathrm{~cm} = 8 \mathrm{~cm} + 150\mathrm{~cm} = 158 \mathrm{~cm}\).

Calculate the change in image distance

Now we can find the change in image distance, \(\Delta d_i\), as the difference between the final image distance and the initial image distance, which is \(\Delta d_i = d_{i2} - d_{i1} = 158 \mathrm{~cm} - 8 \mathrm{~cm} = 150 \mathrm{~cm}\).

Calculate the change in object distance

Let's use the lens formula to find the change in object distance, \(\Delta d_o\). Since both, the initial and final object distances (\(d_{o1}\) and \(d_{o2}\)), use the same eyepiece, the focal length remains the same, \(f = 8 \mathrm{~cm}\). For the initial object distance, we have \[\frac{1}{d_{o1}} = \frac{1}{f} - \frac{1}{d_{i1}} = \frac{1}{8} - \frac{1}{8}\] Hence, \(d_{o1} = \infty\) (i.e., the object is at infinity, as expected for the Sun). For the final object distance, we have \[\frac{1}{d_{o2}} = \frac{1}{f} - \frac{1}{d_{i2}} = \frac{1}{8} - \frac{1}{158}\] We can find \(d_{o2}\) by solving the above equation. Cross-multiplying, we get \[d_{o2} = \frac{158}{\frac{1}{8} - \frac{1}{158}} = 14.65 \mathrm{~cm}\] Now, we can find the change in object distance, \(\Delta d_o\), as the difference between the final object distance and the initial object distance, which is \(\Delta d_o = d_{o2} - d_{o1} = 14.65 \mathrm{~cm} - \infty = -\infty\).

Report the final answer

The eyepiece must be moved 150 cm away from the initial image position (toward the screen) and the object distance should approach infinity (due to the Sun's distance being considered as infinity) to project the Sun's image onto the screen 1.5 meters behind.

Key concepts

The Lens Formula

The lens formula is a fundamental principle used in optics to relate the distances of the object and the image to the focal length of a lens. It is mathematically expressed as:
\[\begin{equation}\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\end{equation}\]In this equation, \(f\) stands for the focal length of the lens, \(d_o\) represents the object distance, and \(d_i\) is the image distance. This formula applies to all situations involving a simple lens, regardless of whether the lens is converging or diverging. The power of this formula lies in its ability to provide a clear relationship between these three critical lens-related distances.

Understanding Focal Length

Focal length, defined as the distance between the lens and its focus, is the key to understanding a lens's optical power. A shorter focal length indicates a stronger lens that can bend light rays more sharply, bringing them to focus in a shorter distance. Conversely, a lens with a long focal length has a weaker optical power and will focus light over a greater distance.
Focal length is important in applications like telescopes and cameras, where the objective is to magnify distant objects or project images. Different focal lengths are required for different applications, with astronomers often preferring long focal lengths to examine celestial objects in great detail.

Calculating Image Distance

The image distance is the measurement from the lens to the point where the image is formed. When using a lens to project an image, such as in telescopes or projectors, calculating the correct image distance is crucial for obtaining a sharp and clear picture.
In some situations, you want to project an image at a particular location — on a screen, for example. The desired image distance can be determined based on the distance you wish to project the image and the known focal length of the lens, using the lens formula.

Role of Image Distance in Projections

Depending on the context, altering the image distance by adjusting the position of a lens or screen results in different magnifications and image qualities. For instance, a larger image distance usually leads to a larger but less bright image.

The Role of Object Distance

In the context of the lens formula, the object distance is the span between the object being viewed and the lens. The object in a telescope's view is typically very far away (the focal point is at infinity), which allows the telescope to bring the light into focus at the focal point of the eyepiece.
However, when projecting an image onto a screen, the situation can be reversed: the screen effectively becomes the 'object' for the viewer, and its distance from the lens influences where the lens needs to be to maintain focus. As the lens moves to project the image, the effective object distance also changes, demonstrating how manipulating one parameter influences the others in accordance with the lens formula.