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Consider a telescope with a small circular aperture of diameter 2.0 centimeters.A) If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resolved if their angular separation is 3.0×10−5 radians?

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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