Respuesta :
Answer:
[tex]4.91803\times 10^{-7}\ m[/tex]
Explanation:
[tex]\theta[/tex] = Angular seperation = [tex]3\times 10^{-5}\ rad[/tex]
[tex]\lambda[/tex] = Wavelength
We have relation
[tex]sin\theta=1.22\dfrac{\lambda}{d}\\\Rightarrow \lambda=\dfrac{dsin\theta}{1.22}\\\Rightarrow \lambda=\dfrac{2\times 10^{-2}sin(3\times 10^{-5})}{1.22}\\\Rightarrow \lambda=4.91803\times 10^{-7}\ m[/tex]
The maximum wavelength of the telescope is [tex]4.91803\times 10^{-7}\ m[/tex]
The maximum wavelength λ at which the two sources can be resolved is of [tex]1.49 \times 10^{-10} \;\rm m[/tex].
Given data:
The diameter of the telescope is, d = 2 cm = 0.02 m.
The angular separation of the two point source of light is, [tex]\theta = 3.0 \times 10^{-5} \;\rm radians = 3.0 \times 10^{-5} \;\rm radians \times \dfrac{\pi}{180}\\\theta = 5.23 \times10^{-7}^{\circ}[/tex].
The angular separation of two point source is also known as apparent distance between the sources, and it is defined as the angle between the two objects as viewed directly by the observer.
The standard relation for the angular separation of telescope is given as,
[tex]sin \theta = 1.22 \times \dfrac{\lambda}{d}[/tex]
here, [tex]\lambda[/tex] is the maximum wavelength of light.
Solving as,
[tex]sin (5.23 \times 10^{-7}) = 1.22 \times \dfrac{\lambda}{0.02}\\\\\lambda = \dfrac{sin (5.23 \times 10^{-7}) \times 0.02}{1.22} \\\\\lambda = 1.49 \times 10^{-10} \;\rm m[/tex]
Thus, we can conclude that the maximum wavelength λ at which the two sources can be resolved is of [tex]1.49 \times 10^{-10} \;\rm m[/tex].
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