The average distance between the earth and mo | vedclass

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(D) The angular resolution of a telescope is given by $	heta = \frac{1.22 \lambda}{D}$,where $\lambda$ is the wavelength and $D$ is the diameter of t

Vedclass · Accepted Answer

$56.51$

Vedclass · Answer

(D) The angular resolution of a telescope is given by $	heta = \frac{1.22 \lambda}{D}$,where $\lambda$ is the wavelength and $D$ is the diameter of the objective lens.Given: $\lambda = 6000 \ \mathring{A} = 6000 	imes 10^{-10} \ m$,$D = 5 \ m$,and distance $R = 38.6 	imes 10^4 \ km = 38.6 	imes 10^7 \ m$.The minimum separation $d$ that can be resolved is $d = R 	imes 	heta$.Substituting the values: $d = R 	imes \frac{1.22 \lambda}{D} = (38.6 	imes 10^7) 	imes \frac{1.22 	imes 6000 	imes 10^{-10}}{5}$.$d = \frac{38.6 	imes 10^7 	imes 1.22 	imes 6 	imes 10^{-7}}{5}$.$d = \frac{38.6 	imes 1.22 	imes 6}{5} = 56.51 \ m$.

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 has a diameter of $5 \ m$ with $\lambda = 6000 \ \mathring{A}$ is ...... $m$.

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The aperture diameter of a telescope is $5\; m$. The separation between the moon and the earth is $4 \times 10^{5} \; km$. With light of wavelength $5500\; \mathring{A}$, the minimum separation between objects on the surface of the moon, so that they are just resolved, is close to......$m$.

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