Question

(a) The planet Pluto and its Moon Charon are separated by \(19,600{\rm{ }}km\). Neglecting atmospheric effects, should the \(5.08 - m\)-diameter Mount Palomar telescope be able to resolve these bodies when they are \(4.50 \times {10^9}{\rm{ }}km\) from Earth? Assume an average wavelength of \(550{\rm{ }}nm\).

(b) In actuality, it is just barely possible to discern that Pluto and Charon are separate bodies using an Earth-based telescope. What are the reasons for this?

Short answer

(a) The telescope is able to resolve the two objects as being separate at a distance of\(s = 5.943 \times {10^5}{\rm{ }}m\).

(b) In actuality, it is just barely possible to discern that Pluto and Charon are separate bodies using an Earth-based telescope. This is because of the limit of the Earth-based telescope.

Step-by-step solution

Concept Introduction

The visible light is a part of the electromagnetic spectrum. The speed of the electromagnetic wave's speed is given by –

\(c = v\lambda \)

Also, the visible light has wavelengths between\(380{\rm{ }}nm\)and\(760{\rm{ }}nm\). The wavelength of the wavelength of electromagnetic wave in a medium is given by –

\({\lambda _n} = \dfrac{\lambda }{n}\)

Where,\(\lambda \)is the wavelength of the electromagnetic wave in vacuum and\(n\)is the redfraction index of the given medium. The thickness of any material is given by –

\(t = \dfrac{\lambda }{n}\)

Young's double slit experiment has unusual results. The light must interact with something small in dimension, such as the closely slits used by Young. To obtain constructive interference for a double slit, the path length difference must be an integral multiple of the wavelength –

\(d(\sin {\rm{ }}\theta ) = m\lambda \)

To obtain destructive interference for a double slit, the path length difference must be a half-integral multiple of the wavelength –

\(d(\sin {\rm{ }}\theta ) = \left( {m + \dfrac{1}{2}} \right)\lambda \)

Where,\(d\)is the distance between the slits,\(\theta \)is the angle from the original direction of the beam and\({\rm{m}}\)is the order of the interference.

The Rayleigh criterion for the difdfraction limit to resolution states that two images are just resolvable whenthe centre of directly over the first minimum ofthe difdfraction pattern of the other. So, the first minimum is at an anglewhich is the angle that separates the two-point objects of –

\(\theta = 1.22\dfrac{\lambda }{D} = \dfrac{x}{d}\)

Where,\(D\)is the aperture's diameter of the lens and\(\theta \)is given by radians. The distance between twoobjects with distance\(r\)is given by –

\(s = r\theta \)

Information Provided

Distance between Pluto and Earthis:

\(\begin{aligned}r = 4.50 \times {10^9}{\rm{ }}km\\ = 4.50 \times {10^9} \times {10^3}{\rm{ }}m\\ = 4.50 \times {10^{12}}{\rm{ }}m\end{aligned}\).

Diameter of telescope lensis: \(D = 5.08{\rm{ }}m\).

Wavelength of the light is:

\(\begin{aligned}\lambda = 550{\rm{ }}nm\\ = \dfrac{{550}}{{{{10}^9}}}\\ = 550 \times {10^{ - 9}}{\rm{ }}m\end{aligned}\).

Calculation for distance

(a) The first minimum angle is represented as –

\(\theta = 1.22\dfrac{\lambda }{D}\)

The distance between two objects with distance\(r\)is given by –

\(\begin{aligned}s &= r\theta \\ &= r1.22\dfrac{\lambda }{D}\\ &= \dfrac{{1.22r\lambda }}{D}\\ &= \dfrac{{1.22 \times 4.50 \times {{10}^{12}}{\rm{ }}m \times 550 \times {{10}^{ - 9}}{\rm{ }}m}}{{5.08{\rm{ }}m}}\\ &= 5.943 \times {10^5}{\rm{ }}m\end{aligned}\)

Therefore, the value for the distance is obtained as \(s = 5.943 \times {10^5}{\rm{ }}m\).

Reasons for Pluto and Charon to be separate

(b) The above value signifies the contribution of difdfraction to the resolvability of a telescope image, when their other factors that can limit or define the resolution of a telescope, for example, difdfraction of the atmosphere, and more significantly the spherical aberration of the telescope, if all the other factors affecting the resolution of the telescope are treated, then the difdfraction limit still remains, in other words, the above-obtained value is only a theoretical maximum, it is the maximum value of a telescope resolution.

Therefore, due to the limit of telescope it is not visible that Pluto and Charon are separate in actual.