The Hubble Space Telescope (Figure 34.33 ) is capable of resolving optical images to an angular resolution of \(2.80 \cdot 10^{-7}\) rad with its 2.40 -m mirror. How large would a radio telescope have to be in order to image an object in the radio spectrum with the same resolution, assuming the wavelength of the waves is \(10.0 \mathrm{~cm} ?\)

To find the wavelength at which the Hubble Space Telescope operates, we need to rearrange the angular resolution formula to solve for \(\lambda\): $$ \lambda = \frac{D \cdot \theta}{1.22} $$ Plug in the given values for the Hubble Space Telescope's angular resolution (\(\theta = 2.80 \cdot 10^{-7}\) rad) and the diameter of its mirror (D = 2.40 m): $$ \lambda_{Hubble} = \frac{2.40 \cdot (2.80 \cdot 10^{-7})}{1.22} $$ Calculate the wavelength: $$ \lambda_{Hubble} = 5.49 \cdot 10^{-7} \mathrm{m} $$

Now we want to find the diameter of the radio telescope to get the same angular resolution as the Hubble Space Telescope at a different wavelength. We can again use the angular resolution formula, rearranging for the diameter: $$ D_{radio} = \frac{1.22 \cdot \lambda_{radio}}{\theta} $$ We are given the wavelength of the radio waves (\(\lambda_{radio} = 10.0 \mathrm{cm} = 0.1 \mathrm{m}\)) and the angular resolution of the Hubble Space Telescope (\(\theta = 2.80 \cdot 10^{-7}\) rad). Plug in the values: $$ D_{radio} = \frac{1.22 \cdot 0.1}{2.80 \cdot 10^{-7}} $$ Calculate the diameter of the radio telescope: $$ D_{radio} = 43.6 \mathrm{m} $$ The radio telescope would have to be approximately 43.6 meters in diameter to achieve the same angular resolution as the Hubble Space Telescope.