The telescope at a small observatory has objective and...

The telescope at a small observatory has objective and eyepiece focal lengths respectively of 19.2 m and 12 cm. If this telescope is used to view a 1700 m diameter lunar crater on the surface of the moon 3.77 × 10^8 m from the surface of the Earth, determine the following. (a) angular magnification of the telescope (b) size of the first image m (c) length of the barrel of the telescope m (d) angle subtended at the unaided eye by the lunar crater rad (e) angle subtended at the eye when the lunar crater is viewed through the telescope rad

Transcript

00:01 Hello everyone.

00:01 So here we are given that there is a telescope at small observatory.

00:06 The focal length of the objective piece is 19 .2 meter and the eye piece is 12 centimeter.

00:14 We are given that the diameter of the lunar crater is 1700 meters and distance of moon is 3 .17 and 10 .8 meters.

00:26 We have to find the optical magnification, the size of the image.

00:35 Then we have to find the length of the barrel and then we have to find the angle subtended at the uneated eye by the lunar crater and then if seen from the telescope.

00:46 So let us start with part a.

00:50 We know that optical magnification is equal to the ratio of the focal length of objector to that of ips.

00:58 So this is equal to 19 .2 meters divided by 12 centimetre that is 12 multiplied by 10 .5.

01:06 Minus 2.

01:08 So this is equal to 1 .6 multiplied by 10 raised to the power 2.

01:15 For simply magnification is equal to 160.

01:19 This is all about part a.

01:21 Now let us move out to part b.

01:23 We have the size of the image.

01:28 The size of the image is given by the angle subtended by the telescope multiplied by the focal length.

01:38 So we have the size of the image.

01:39 So we have a have angle subtended is equal to the diameter of the object that is the lunar crater here divided by the distance from telescope and this multiplied by the focal length of the objective.

02:04 So let us start by mutting the values.

02:08 We have 1700 divided by 3 .77 multiplied by 10 raised to the power 8 multiplied by 19 .2.

02:22 So this is equal to we have 8 ,657 .8 multiplied by 10 raised to the power minus 8.

02:34 Or simply, size is equal to 8 .66 multiplied by 10 raised to the power minus 5 meters...

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