The distance of the moon from earth is 3.8 ti | vedclass

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$(A)$ The limit of resolution of a telescope is given by $\Delta 	heta = \frac{1.22 \lambda}{d}$, where $\lambda$ is the wavelength of light and $d$

Vedclass · Accepted Answer

$51$

Vedclass · Answer

$(A)$ The limit of resolution of a telescope is given by $\Delta 	heta = \frac{1.22 \lambda}{d}$, where $\lambda$ is the wavelength of light and $d$ is the diameter of the telescope aperture.Given: $\lambda = 5500 	imes 10^{-10} 	ext{ m}$, $d = 500 	ext{ cm} = 5 	ext{ m}$, and $r = 3.8 	imes 10^5 	ext{ km} = 3.8 	imes 10^8 	ext{ m}$.The separation $x$ between two points that can be resolved is given by $x = r \Delta 	heta$.Substituting the values: $x = \frac{1.22 \lambda r}{d}$.$x = \frac{1.22 	imes (5500 	imes 10^{-10} 	ext{ m}) 	imes (3.8 	imes 10^8 	ext{ m})}{5 	ext{ m}}$.$x = \frac{1.22 	imes 5.5 	imes 10^{-7} 	imes 3.8 	imes 10^8}{5} 	ext{ m}$.$x = \frac{1.22 	imes 5.5 	imes 3.8 	imes 10}{5} 	ext{ m} = 51 	ext{ m}$.

The distance of the moon from earth is $3.8 \times 10^5 \text{ km}$. The eye is most sensitive to light of wavelength $5500 \text{ Å}$. The separation of two points on the moon that can be resolved by a $500 \text{ cm}$ telescope will be......$m$.

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