Animated Solution for Physics - Optics: The aperture of a telescope is $5\text{ m}$. The separation between the moon and the earth is $4\times 10^5\text{ km}$. With light of wavelength of $5500\text{ A˚}$, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to

Have you ever looked up at the night sky and wondered what it would take to see the footprints left by the Apollo astronauts on the Moon? While a backyard telescope might show you craters and maria, resolving tiny details requires massive instruments. This problem takes us on a journey to understand the fundamental limits of optical resolution, governed by the wave nature of light itself.

Imagine standing on Earth, peering through a telescope with a massive $5\text{ m}$ aperture. You are looking at the Moon, which is located at a distance $d=4\times 10^5\text{ km}$ away. On the surface of the Moon, there are two distinct objects, separated by a linear distance $D$. Our goal is to find the minimum value of $D$ such that our telescope can just barely tell them apart as two separate entities, rather than a single blurred blob.

When light from these two objects enters our telescope, it undergoes diffraction. The circular aperture of the telescope creates a diffraction pattern for each object, consisting of a central bright disk (the Airy disk) surrounded by fainter rings. According to Rayleigh's Criterion, two point sources are just resolved when the center of the diffraction pattern of one source falls exactly on the first minimum of the diffraction pattern of the other.

Mathematically, this gives us the minimum angular separation ${\theta }_R$:

where $\lambda$ is the wavelength of the light and $a$ is the diameter of the aperture.

Now, let's connect this angular separation to the physical reality on the Moon. The two objects and the telescope form a long, skinny triangle. Because the distance to the Moon is so vast compared to the separation between the objects, the angle ${\theta }_R$ is incredibly small. In this small-angle regime, we can use the arc length formula:

Substituting our expression for ${\theta }_R$, we arrive at our master equation:

Before we plug in the numbers, we must ensure absolute dimensional consistency. Physics is unforgiving when it comes to mixed units! Let's convert everything to standard SI units (meters).

The aperture is already in meters: $a=5\text{ m}$.

The distance to the Moon is given in kilometers: $d=4\times 10^5\text{ km}=4\times 10^8\text{ m}$.

The wavelength is given in Angstroms: $\lambda =5500\text{ A˚}=5500\times 10^{-10}\text{ m}$.

Now, we substitute these pristine values into our master equation:

Let's handle the powers of 10 first to simplify the landscape. We have $10^{-10}$ and $10^8$, which combine to give $10^{-2}$.

Multiplying $5500$ by $10^{-2}$ simply shifts the decimal point two places to the left, leaving us with $55$.

This is much less intimidating! We can divide $55$ by $5$ to get $11$.

So, the absolute minimum separation between the two objects for them to be resolved by this telescope is approximately $54\text{ m}$.

Looking at our options, we don't see $54\text{ m}$ exactly. However, the question asks for the value that is close to our result. The options are $600\text{ m}$, $20\text{ m}$, $60\text{ m}$, and $200\text{ m}$. The value $60\text{ m}$ is by far the closest to our calculated $53.68\text{ m}$. In many physics problems, especially those involving astronomical distances and order-of-magnitude estimations, finding the closest matching option is the intended path.