At Kavalur in India, the astronomers using a telescope whose objective had a diameter of one meter started using a telescope of diameter $$2.54\ m$$. This resulted in
The resolving power of a telescope is given by,
$$R_p=\dfrac{D}{1.22\times \lambda}$$ where, $$D$$ is the diameter of the objective lens and $$\lambda $$ is wavelength of the light.
Hence, we can conclude that, $$R_p \propto D$$.
$$\therefore \dfrac{R_{p_1}}{R_{p_2}}=\dfrac{D_1}{D_2}=\dfrac{1}{2.54}$$
or, $$R_{p_2}=2.54 \times R_{p_1}$$
Hence, option (A) is the correct answer.
An astronomical telescope has an angular magnification of magnitude $$5$$ for distant object. The separation between the objective and the eye piece is $$36\ cm$$and final image is formed at infinity. The focal length of the objective and focal length of eye-piece respectively are:
The large aperture of telescope are used for
The minimum magnifying power of telescope is $$M$$. If the focal length of its eye lens is halved, the magnifying power will become:-
A telescope using light having wavelength $$5000A^o$$ and using lenses of focal $$2.5$$ and $$30cm$$. If the diameter of the aperture of the objective is $$10cm$$, then the resolving limit and magnifying power of the telescope is respectively
If the telescope is reversed i.e. seen from the objective side
In an astronomical telescope, the focal lengths of two lenses are $$180\ cm$$ and $$6\ cm$$ respectively. In normal adjustment, the magnifying power will be
All of the following statements are correct except
When diameter of the aperture of the objective of an astronomical telescope if increased, its
The length of a telescope is $$36\ cm$$. The focal lengths of its lenses can be
The diameter of objective of a telescope is $$1\ m$$. Its resolving limit for the light of wave length $$4538\ A^o$$, will be