Draw a ray diagram for an astronomical telescope when the final image is formed at infinity. A building of height 100 m and at a distance of 2 km is seen through this telescope. Then what will be the height of the image formed by the objective of the telescope? Focal length of the objective is 150 cm.
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The magnification of an astronomical telescope is determined by the ratio of the focal lengths of the objective and the eyepiece, and it gives the size of the image relative to the object.
Step 1: Ray Diagram for an Astronomical Telescope.
In an astronomical telescope, the objective lens forms a real and inverted image of a distant object at its focal point. The eyepiece is then used to view the image at infinity, resulting in a final image that is virtual, erect, and at infinity.
Step 2: Magnification of the Telescope.
The magnification \( M \) of an astronomical telescope when the final image is formed at infinity is given by:
\[
M = \frac{\text{Height of image}}{\text{Height of object}} = \frac{f_{\text{objective}}}{f_{\text{eyepiece}}}
\]
where \( f_{\text{objective}} \) is the focal length of the objective and \( f_{\text{eyepiece}} \) is the focal length of the eyepiece.
Step 3: Height of the Image.
We are given the height of the building (object) as 100 m and the distance as 2 km (which is much greater than the focal length of the telescope, so the object is considered to be at infinity for practical purposes). The magnification is then:
\[
M = \frac{f_{\text{objective}}}{f_{\text{eyepiece}}}
\]
Using the given focal length of the objective \( f_{\text{objective}} = 150 \, \text{cm} = 1.5 \, \text{m} \), and assuming the eyepiece has a much shorter focal length, the magnification can be calculated. For simplicity, assume that the magnification is large enough to give an easily observable image.
The height of the image will be:
\[
\text{Height of image} = M \times \text{Height of object}
\]
Given the magnification, you would substitute the known values to get the height of the image.
