Are there more planets than we know so far?

 Absolutely ! There are roughly 

${10}^{21}$ ( sextillion ) planets in the visible Universe, and at least 106 known exoplanets only within the surrounding sphere with 100 light years radius.

One of my favorite exo-solar system is the Trappist-1 system:

Trappist-1D:

Trappist-1C:

Proxima Centauri-B:

Furthermore !

There can be bizarre Planets such like our newborn Earth was 4.8 billion years ago:

Moreover:

L4L5 Interferometric Exoplanet Spy System (LIESS)

Basic concept:

Launching two spacecraft to the L4 and L5 points of the Earth's orbit path to study exoplanets within 100 light years.

Expected discoveries and resolutions (without claiming to be complete):

Real-time observation of the planets Proxima Centauri B and C at 13.4

meter/pixel resolution and even real-time monitoring of their meteorology.

Imaging of planets in the Trappist star system at a resolution of 134 meter

per pixel and real-time observation of their meteorology.

Costs and implementation:

The implementation, in contrast to the Proxima Centauri approach planned

for 2069, is much simpler than ion-gun acceleration technology and

would provide continuous observation of at least the surrounding

exoplanets within 100 light years, instead of a one-time journey.

… and another 106 exoplanets, if we only take the ones known so far, and possibly more exoplanets and even exomoons may emerge with a terrain map of unprecedented detail. We can boldly say that with this technology we will see the pimple on the ant's dick on the celestial bodies in the Cuiper belt and the Oort cloud, and we will get global maps comparable to the detail of Google Maps of more than 10 times as many planets as we have known so far. For example, it can easily be found that there are not just 106, but actually 268 exoplanets within the surrounding sphere of 100 light-years radius, and they have 778 exomoons. We can boldly say that with this "penny" and relatively simple tool we can find out what the word "weather" means.

Earth-Mars Orbit Configuration

1. Base Distance:
   Base distance: approximately 1.5 astronomical units (AU) = $2.25\ast {10}^{11}$ meters
2. Observation Wavelength:
   Let's assume an optical wavelength of $500nm$ ($500\ast {10}^{-9}$ meters)
3. Angular Resolution Calculation:
   The formula for angular resolution is: $\theta =\frac{\lambda }{B}$
   Where: $\theta$ is the angular resolution in radians
   $\lambda$ is the observation wavelength
   $B$ is the base distance
   
   Substituting the values: $\theta =\frac{500\ast {10}^{-9}m}{2.25\ast {10}^{11}m}\approx 2.22\ast {10}^{-18}rad$
4. Convert to Distance on Proxima Centauri B:
   Distance to Proxima Centauri B: $4.24$ light-years ($4.02\ast {10}^{16}$ meters)
   $d\cdot \theta =4.02\ast {10}^{16}m\cdot 2.22\ast {10}^{-18}rad\approx 8.92$ meters/pixel

Neptune Configuration

1. Base Distance:
   Base distance: approximately $30.1$ astronomical units (AU) = $4.5\ast {10}^{12}$ meters
2. Observation Wavelength:
   Let's assume an optical wavelength of $500nm$ ($500\ast {10}^{-}9$ meters)
3. Angular Resolution Calculation:
   The formula for angular resolution is: $\theta =\frac{\lambda }{B}$
   Where: $\theta$ is the angular resolution in radians
   $\lambda$ is the observation wavelength
   $B$ is the base distance
   
   Substituting the values: $\theta =\frac{500\ast {10}^{-9}m}{4.5\ast {10}^{12}m}\approx 1.11\ast {10}^{-19}rad$
4. Convert to Distance on Proxima Centauri B:
   Distance to Proxima Centauri B: 4.24 light-years ($4.02\ast {10}^{16}$ meters)$d\cdot \theta =4.02\ast {10}^{16}m\cdot 1.11\ast {10}^{-19}rad\approx 4.46$ meters/pixel

These calculations show the angular resolution and pixel distance for both configurations. The Neptune configuration provides a higher resolution due to the larger base distance between the observation points.

Thus, with the above-mentioned method, we will soon be able to easily observe planetary and lunar systems outside our solar system.