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The magnification of the telescope image is

(focal length of the objective) divided by (focal length of the eyepiece).

The focal length of the objective is fixed.
Decreasing the focal length of the eyepiece increases the magnification of the image.
(But it also makes the image dimmer.)

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Waldo Mayer

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3y ago

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What is a barlow lens on a telescope?

A Barlow lens is an accessory used in telescopes to increase the focal length, resulting in magnification of the image. It allows the telescope to achieve higher magnification without needing to switch to a higher power eyepiece. By inserting the Barlow lens between the telescope and eyepiece, it effectively doubles or triples the focal length of the telescope.


What will decreasing a telescope's eyepiece focal length?

The magnification of the telescope image is(focal length of the objective) divided by (focal length of the eyepiece).The focal length of the objective is fixed.Decreasing the focal length of the eyepiece increases the magnification of the image.(But it also makes the image dimmer.)


What happens to the active force with a decrease in the muscle length?

It decreaseses.


How can the magnifying power of a telescope be increased?

a telescope's magnification is calculated as the ratio of the focal length of the primary objective to the focal length of the eyepiece. Since a telescope is defined by the primary objective, this part of it is essentially unchangeable. Therefore, the way to increase magnification is to decrease the focal length of the eyepiece. For example, a 1000mm objective and a 25mm eyepiece yields a magnification of (1000/25) 40x. Changing the eyepiece to a 10mm eyepiece increases magnification to (1000/10) 100x.


Telescope magnification power with a 30 mm eyepiece for a telescope?

To calculate the magnification power with a 30 mm eyepiece, you need to divide the focal length of the telescope by the focal length of the eyepiece. For example, if the telescope has a focal length of 600 mm and you use a 30 mm eyepiece, the magnification would be 20x (600 mm / 30 mm = 20x).


If the primary mirror of a telescope have a focal length of 225 centimeters and the eyepiece has a focal length of 7.5 millimeters then what is the magnifying power of the telescope?

The magnifying power of a telescope is the focal length of the scope in millimeters, divided by the focal length of the eyepiece in millimeters. Focal length of scope: 225cm=2250mm Focal length of eyepiece: 7.5mm 2250/7.5= 300X


As the speed and length of wave decrease what happens to the wave?

As the speed and length of a wave decreases, the frequency of the wave remains constant. This means that the wave will experience a decrease in wavelength, which is inversely proportional to the decrease in speed. The energy of the wave will also decrease.


How does the eyepiece affect the total magnifacation in a telescope?

A telescope consists of two lenses. 1) The main lens which collects the light ( it is relatively bigger that eyepiece). 2) Eye piece , through which we see. Magnification of a telescope depends on the focal length of the eye piece and the main lens. Magnification = Focal length of the main lens / Focal length of the eyepiece . For example : If the focal length of the main lens is 12 units and the focal length of the eyepiece is 2 units , then the magnification will be 12/2 = 6.When the focal length of the main lens is constant , the focal length of the eyepiece is inversely proportional to the magnification.


What is the power of the eyepiece?

The magnification, or power, at which a telescope is operating is a function of the focal length of the telescope's main (objective) lens (or primary mirror) and the focal length of the eyepiece employed.


What would be the magnification of a telescope with a focal length of 3000 millimeters using an eyepiece of focal length equal to 15 mm?

The magnification of a telescope is calculated by dividing the focal length of the telescope by the focal length of the eyepiece. In this case, the magnification would be 3000 mm (telescope focal length) divided by 15 mm (eyepiece focal length), which equals a magnification of 200x.


As the speed and length of a wave decrease what happens to the wave?

As the speed and length of a wave decrease, the frequency remains the same while the wavelength decreases. This means that the wave's energy decreases as well.


How do you calculate magnification?

Divide the focal length of the objective lens by the focal length of the eyepiece.