Tag: observing space: telescopes

In a terrestrial telescope the focal length erecting lens is $20cm$. The length of the telescope $96cm$ . If the magnifying power of the telescope $10$. Then the focal length of eye -piece and objective are respectively 

The aperture of the largest telescope in the world is $5 m,$ if the separation between the Moon and the Earth is $4 \times 10^5 km$ and the wavelength of the visible light is $5000 \overset {o}{A}$ then the minimum separation between the objects on the surface of the Moon which can be just resolve is approximately

If an astronomical telescope has objective and eye-pieces of focal length 200 cm and 4 cm respectively,then the magnifying power of the telescope for the normal vision is:

Focal lengths of the objective lens and eye-piece of an astronomical telescope are $2m$ and $0.05m$. Find the length of telescope in normal adjustment.

The diameter of moon is $3.5\times{10}^{3}km$ and its distance from the earth is $3.8\times{10}^{5}km$. The focal length of the objective and eyepiece are $4m$ and $10cm$ respectively. The angle subtended by the diameter of the image of the moon will be approximately

The magnifying power an astronomical telescope for normal adjustment is -

If for a given telescope $D = 22 \,mm$ and $\lambda = 6 \times 10^{-7}m$, then the minimum value of the angle subtended by two stars that could be resolved is approximately:

The focal length of objective of an astronomical telescope is $1m$. If the magnifying power of telescope is $8$, then what is length of telescope for relaxed eye?

The diameter of the lens of a telescope is 1.22 m., the wavelength of light is $5000{A^0}$ the resolution power of the telescope is 

If an object subtend angle of $2^0$ at eye when seen through telescope having objective and eyepiece of focal length $f _0=60cm$ and $f _e=5cm$ respectively than angle subtend by image at eye piece will be