Question

A telescope allows one to view large objects that are far away. Refractive telescopes function by having the objective form a reduced real image (I) of the object, while the eyepiece forms a virtual, enlarged image of the object. The objects astronomers observe through refracting telescopes are at such great distances that the first image, I, is very nearly at a focal point of the objective. The final image, I , is formed at infinity, when the first image is at a focal point of the eyepiece. The length of the telescope, defined as the distance between the objective and the eyepiece, is the sum of the focal lengths of the objective and the eyepiece, $f_1+f_2$. Figure 1 can't copy The magnitude of the telescope's angular magnification, M , is equal to the ratio of the focal length of the objective to the focal length of the eyepiece. The calculation of M is shown below: $$ M=\frac{\theta^{\prime}}{\theta}, $$ where $\theta=-y^{\prime} / f_1$ and $\theta^{\prime}=y^{\prime} / f_2$. After substituting for $\theta$ and $\theta^{\prime}$ : $$ \mathrm{M}=-\frac{\mathrm{y}^{\prime} / f_2}{\mathrm{y}^{\prime} / f_1}=-\frac{f_1}{f_2} $$ In reflecting telescopes, the other broad category of telescopes, a large concave mirror replaces the comparatively smaller objective of the refracting telescope. The mirror reflects the image to a smaller plane mirror, which reflects the image to the eyepiece. There are three types of focus used in reflecting telescopes: prime focus, Newtonian focus, and Cassegrainian focus. In telescopes with prime focus, there is no second plane mirror, and the eyepiece actually sits in the path of the light. The eyepiece is arranged differently with respect to the mirrors. The focal length of a spherical mirror depends on the: A. radius of curvature of the mirror. B. index of refraction of the mirror. C. absorbance of the mirror. D. intensity of the incident light.

   A telescope allows one to view large objects that are far away. Refractive telescopes function by having the objective form a reduced real image (I) of the object, while the eyepiece forms a virtual, enlarged image of the object. The objects astronomers observe through refracting telescopes are at such great distances that the first image, I, is very nearly at a focal point of the objective. The final image, I , is formed at infinity, when the first image is at a focal point of the eyepiece. The length of the telescope, defined as the distance between the objective and the eyepiece, is the sum of the focal lengths of the objective and the eyepiece, $f_1+f_2$.
Figure 1 can't copy
The magnitude of the telescope's angular magnification, M , is equal to the ratio of the focal length of the objective to the focal length of the eyepiece. The calculation of M is shown below:

$$
M=\frac{\theta^{\prime}}{\theta},
$$

where $\theta=-y^{\prime} / f_1$ and $\theta^{\prime}=y^{\prime} / f_2$. After substituting for $\theta$ and $\theta^{\prime}$ :

$$
\mathrm{M}=-\frac{\mathrm{y}^{\prime} / f_2}{\mathrm{y}^{\prime} / f_1}=-\frac{f_1}{f_2}
$$


In reflecting telescopes, the other broad category of telescopes, a large concave mirror replaces the comparatively smaller objective of the refracting telescope. The mirror reflects the image to a smaller plane mirror, which reflects the image to the eyepiece. There are three types of focus used in reflecting telescopes: prime focus, Newtonian focus, and Cassegrainian focus. In telescopes with prime focus, there is no second plane mirror, and the eyepiece actually sits in the path of the light. The eyepiece is arranged differently with respect to the mirrors.

The focal length of a spherical mirror depends on the:
A. radius of curvature of the mirror.
B. index of refraction of the mirror.
C. absorbance of the mirror.
D. intensity of the incident light.
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MCAT: The Berkley Review Physics Book II
MCAT: The Berkley Review Physics Book II
kalbaba 1st Edition
Chapter 10, Problem 60 ↓
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A telescope allows one to view large objects that are far away. Refractive telescopes function by having the objective form a reduced real image (I) of the object, while the eyepiece forms a virtual, enlarged image of the object. The objects astronomers observe through refracting telescopes are at such great distances that the first image, I, is very nearly at a focal point of the objective. The final image, I , is formed at infinity, when the first image is at a focal point of the eyepiece. The length of the telescope, defined as the distance between the objective and the eyepiece, is the sum of the focal lengths of the objective and the eyepiece, $f_1+f_2$. Figure 1 can't copy The magnitude of the telescope's angular magnification, M , is equal to the ratio of the focal length of the objective to the focal length of the eyepiece. The calculation of M is shown below: $$ M=\frac{\theta^{\prime}}{\theta}, $$ where $\theta=-y^{\prime} / f_1$ and $\theta^{\prime}=y^{\prime} / f_2$. After substituting for $\theta$ and $\theta^{\prime}$ : $$ \mathrm{M}=-\frac{\mathrm{y}^{\prime} / f_2}{\mathrm{y}^{\prime} / f_1}=-\frac{f_1}{f_2} $$ In reflecting telescopes, the other broad category of telescopes, a large concave mirror replaces the comparatively smaller objective of the refracting telescope. The mirror reflects the image to a smaller plane mirror, which reflects the image to the eyepiece. There are three types of focus used in reflecting telescopes: prime focus, Newtonian focus, and Cassegrainian focus. In telescopes with prime focus, there is no second plane mirror, and the eyepiece actually sits in the path of the light. The eyepiece is arranged differently with respect to the mirrors. The focal length of a spherical mirror depends on the: A. radius of curvature of the mirror. B. index of refraction of the mirror. C. absorbance of the mirror. D. intensity of the incident light.
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Transcript

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00:01 For this problem on the topic of geometrical optics, for a given telescope, we want to find the magnification of the 36 inch refractor.
00:11 And if we are told the focal length of the objective lens is 17 .37 meters, and the focal length of the eye piece is 22 millimeters.
00:19 We then want to explain the significance of the negative sign in the magnification equation.
00:25 Now, the focal lens of the objective and eyepiece in the lick refractor in the lick observatory are given as.
00:31 As fo, which is 17 .37 meters, and fe equal to 0 .022 meters, respectively...
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