Problem 95 - Physics

Question

The telescope at Yerkes Observatory in Wisconsin has an objective whose focal length is \(19.4 \mathrm{~m} .\) Its eyepiece has a focal length of \(10.0 \mathrm{~cm} .\) (a) What is the angular magnification of the telescope? (b) If the telescope is used to look at a lunar crater whose diameter is \(1500 \mathrm{~m},\) what is the size of the first image, assuming the surface of the moon is \(3.77 \times 10^{8} \mathrm{~m}\) from the surface of the earth? (c) How close does the crater appear to be when seen through the telescope?

Step-by-Step Solution

Answer

(a) 194, (b) \(7.72 \times 10^{-5}\) m, (c) 19,442 km.

Step 1: Understand 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 lens and \( f_e \) is the focal length of the eyepiece lens. Here, \( f_o = 19.4 \) m and \( f_e = 10.0 \) cm, which needs to be converted to meters first.

Step 2: Convert Units for Eyepiece Focal Length

Convert the focal length of the eyepiece from centimeters to meters: \( 10.0 \) cm = \( 0.1 \) m. This is necessary to maintain consistent units when calculating the angular magnification.

Step 3: Calculate Angular Magnification

Using the formula from Step 1, substitute the converted focal lengths: \( M = \frac{19.4}{0.1} = 194 \). Thus, the angular magnification of the telescope is 194.

Step 4: Calculate Size of First Image

To find the size of the first image \( h' \), use the formula \( h' = \frac{h_o imes f_o}{d_o} \) where \( h_o \) is the diameter of the crater (1500 m), \( f_o \) is 19.4 m, and \( d_o \) is the distance to the moon (\( 3.77 \times 10^{8} \) m).

Step 5: Compute Size of First Image

Substitute the known values into the equation: \( h' = \frac{1500 \times 19.4}{3.77 \times 10^8} \) which simplifies to \( h' = 7.72 \times 10^{-5} \) m. So, the size of the first image is \( 7.72 \times 10^{-5} \) m.

Step 6: Determine Apparent Distance

The apparent distance \( d' \) when viewed through the telescope can be found using the formula \( d' = \frac{d_o}{M} \). Use \( d_o = 3.77 \times 10^8 \) m and \( M = 194 \).

Step 7: Calculate Apparent Distance

Plug in the values: \( d' = \frac{3.77 \times 10^8}{194} \approx 19.44 \times 10^6 \) m or approximately 19,442 km. Thus, the distance appears nearly 19,442 km due to magnification.

Key Concepts

Telescope

A telescope is a fascinating optical instrument that allows us to observe distant objects with enhanced detail. At its core, a telescope typically consists of an objective lens or mirror and an eyepiece. These components work together to collect and focus light. - **Objective Lens/Mirror**: This is the larger lens or mirror in a telescope. It gathers light from distant objects and brings it to a focal point. - **Eyepiece**: Located where the observer looks through the telescope, the eyepiece magnifies the image formed by the objective. The purpose of a telescope is to increase the angular size and brightness of distant objects, making details sharper and clearer. For instance, the telescope used in Yerkes Observatory has a significant objective focal length, which contributes to its high angular magnification capability. This means it can make distant objects, like lunar craters, easier to see.

Focal Length

Focal length is a crucial aspect of how a telescope functions. It is the distance from the lens (or mirror) to the point where light converges to a focus. - A longer focal length in a telescope’s objective allows for greater detail and magnification of distant objects. - By comparison, a shorter focal length in the eyepiece indicates a higher magnification. To find the angular magnification of a telescope, you divide the focal length of the objective by the focal length of the eyepiece. In the exercise, converting measurements to consistent units is vital. The significant focal length of 19.4 meters for the Yerkes Observatory telescope’s objective, compared to the eyepiece’s 0.1 meter, yields a magnification of 194. This high magnification makes the telescope powerful for detailed astronomical observation.

Lunar Crater

Lunar craters, which are depressions on the moon's surface created by impacts, are excellent subjects for telescopic observation. These features are stunning to view due to their varied sizes and complex structures. - Observing a lunar crater requires a telescope with sufficient magnification and resolution to differentiate the fine details of its rim and floor. - Calculating the apparent size and distance of a lunar crater, like the one in the Yerkes Observatory problem, is a means of leveraging the telescope’s capabilities to translate vast distances into viewable images. For instance, using the telescope's magnification, a crater appearing 19,442 km away is much closer than its actual 377,000 km distance. This difference allows for a detailed exploration of lunar topography, making a crater look closer and larger, enhancing the observation experience.