Question

A telescope is constructed from two lenses with focal lengths of 95.0 cm and 15.0 cm, the 95.0-cm lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 m tall, 3.00 km away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

Short answer

(a) The angular magnification is 6.33. (b) The image height is 0.0190 m. (c) The angular size of the final image is 0.126 radians.

Step-by-step solution

Understanding the Problem

We have two lenses forming a telescope. The first lens is the objective with a focal length of 95.0 cm, and the second is the eyepiece with a focal length of 15.0 cm. We need to find the angular magnification, the height of the image for a specific object, and the angular size of the final image.

Finding Angular Magnification

The angular magnification \(M\) of a telescope is given by the formula: \[ M = \frac{f_o}{f_e} \] Where \(f_o\) is the focal length of the objective, and \(f_e\) is the focal length of the eyepiece. Substitute \(f_o = 95.0\) cm and \(f_e = 15.0\) cm: \[ M = \frac{95.0}{15.0} = 6.33 \] Thus, the angular magnification is 6.33.

Calculating the Image Height

For the objective lens, we use the lens formula: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] Since the object and image are at infinity, the image formed by the objective is real and at its focal point. Therefore, \(d_i = f_o = 95.0\) cm = 0.95 m. We can find the image height using magnification \(m\) formula: \[ m = -\frac{d_i}{d_o} \] Since the object height \(h_o = 60.0\) m, and \(d_o = 3000\) m: \[ m = -\frac{0.95}{3000} \approx -0.000317 \] Therefore, the image height \(h_i\) is: \[ h_i = m \times h_o = -0.000317 \times 60.0 = -0.0190 \] The height of the image formed by the objective is approximately \( 0.0190 \) m.

Determining Angular Size of the Final Image

The angular size \(\theta\) of the final image for the eyepiece, when the image is at infinity, is approximately equal to the angular magnification multiplication with a small angle subtended by the object directly. Since the object viewed was at an extremely large distance, consider: \[ \theta = M \times \text{angle subtended by the actual height} \] Since both object and final image are at infinity with small angles, we approximate angular size using magnification: \[ \theta \approx \frac{h_i}{f_o} \times M = \frac{0.019}{0.95} \times 6.33 \] \[ \theta = 0.126 \text{ rad} \] The angular size of the final image is approximately 0.126 radians.

Key concepts

Objective Lens Focal Length

The objective lens is an essential part of any telescope. Its primary role is to gather light from distant objects and converge it to form a real image. In our case, the objective lens has a focal length of 95.0 cm. This measurement tells us how far from the lens the incoming parallel light rays converge to a point.

A longer focal length, like the one we have here, implies that the telescope can focus light from distant objects, which is ideal for astronomical observations. The objective lens with this focal length will produce a real image of the object at its focal plane, laying the groundwork for the eyepiece to magnify this image.

• Objective lens gathers light
• Has a focal length of 95.0 cm
• Converges light to form an image

Eyepiece Lens Focal Length

The eyepiece lens serves as a magnifier for the image created by the objective lens. With a focal length of 15.0 cm, the eyepiece manages to spread out the final light rays slightly, ensuring that a larger angular image reaches the observer's eye.

The purpose of a shorter eyepiece focal length is to amplify the angular magnification of the telescope. In simpler terms, the shorter the focal length of the eyepiece, the more it boosts the size of the image seen through the telescope, making distant objects appear closer and more detailed.

• Eyepiece magnifies the image
• Focal length is 15.0 cm
• Creates larger angular image

Image Height Calculation

Calculating the image height involves understanding the concept of image formation through lenses. In this setup, the objective lens forms an image of a distant object, such as a building 60.0 m tall located 3.00 km away, at its focal point. Since both the object and the image are at infinity for practical purposes, we assume that the image is formed at the focal length of the objective lens.

Using the lens formula, which relates the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\), we find that for objects at such large distances, \(d_i\) approximately equals \(f_o\). As a result, we determine the image height \(h_i\) by applying the magnification formula. This height conveys how large the image of the building is seen through the telescope, allowing for further analysis.

• Objective lens forms image at focal point
• Image height relates to object height and distance
• For distant objects, \(d_i ≈ f_o\)