If the diameter of a radar dish is doubled, what happens to its resolving power assuming that all other factors remain unchanged?

      θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

Explanation:

The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.

The first minimum occurs for m = 1, so the diffraction equation of a slit remains

        a sin θ = λ

in general, the diffraction patterns occur at very small angles, so

        sin θ = θ

          θ = λ / a

in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.

        θ = 1.22 λ /a

In this exercise we are told that the opening changes

         a’ = 2 a

we substitute

          θ ‘= 1.22  λ / 2a

          θ' = (1.22 λ / a) 1/2

          θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

the diagram which shows forces as either balanced = same speed or unbalanced = accelerarting

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