The resolving power of a telescope whose lens has a diameter of $$1.22\ m$$ for a wavelength of $$5000\ A^o$$ is
Resolving power $$=\dfrac{d}{1.22\ \lambda}=\dfrac{1.22}{1.22 \times 5000\times 10^{-10}}=2\times 10^6$$
Two stars are 10 light years away from the earth. They are seen through a telescope of objective diameter 30 cm. The wavelength of light is 600 nm. To see the stars just resolved by the telescope, the minimum distance between them should be (1 light year $$= 9.46 \times 10^{15}m$$) of the order of:
The angular resolution of telescope of 10 cm diameter at a wavelength of 5000 $${ A }^{ \circ }$$ is of the order of :
If $$F_o$$ and $$F_e$$ are the focal length of the objective and eye-piece respectively of a telescope, then its magnifying power will be
The diameter of the objective of the telescope is $$0.1$$ metre and wavelength of light is $$6000\ A^o$$. Its resolving power would be approximately
The average distance between the earth and moon is $$38.6\times 10^4 km$$. The minimum separation between the two points on the surface of the moon that can be resolved by a telescope whose objective lens diameter of $$5\ m$$ with $$\lambda =6000\overset {o}{A}$$ is
The distance of the moon from earth is $$3.8\times 10^5 km$$. The eye is most sensitive to light of wavelength $$5500\ \overset{o}{A}$$. The separation of two points on the moon that can be resolved by a $$500\ cm$$ telescope will be
Two convex lenses of same focal length but of aperture $$ A1$$ and $$ A2 ( A2 < A1)$$ , are used as the objective lenses in two astronomical telescope having identical eyepieces, what is the ratio of there resolving power? Which telescope will you prefer and why? Give reason.