The modulation index is the ratio of the message amplitude to the carrier amplitude, and both can be read off the envelope — giving option 3:

\(m=\dfrac{V_{max}-V_{min}}{V_{max}+V_{min}}\)

The derivation from the envelope. An AM wave's envelope swings between

\(V_{max}=V_{c}+V_{m},\qquad V_{min}=V_{c}-V_{m}\)

Adding and subtracting these two recovers the individual amplitudes:

\(V_{max}+V_{min}=2V_{c},\qquad V_{max}-V_{min}=2V_{m}\)

and dividing one by the other gives

\(\dfrac{V_{max}-V_{min}}{V_{max}+V_{min}}=\dfrac{2V_{m}}{2V_{c}}=\dfrac{V_{m}}{V_{c}}=m\)

Why the form must be a ratio of a difference to a sum. Two independent checks settle it without the algebra.

Dimensions. The index is a pure number, so both numerator and denominator must be voltages. Options 1 and 2 square one of them, giving units of volts and reciprocal volts respectively — neither is dimensionless, so both are impossible.

Range. The index must satisfy \(0\le m\le1\) for undistorted AM. In option 3, since \(V_{min}\ge0\), the numerator can never exceed the denominator, so m is automatically bounded by 1 — exactly right. Option 4 is its reciprocal and is therefore always at least 1, which no valid index can be.

Reading the two extremes off the formula.

Why this particular form is the useful one. \(V_{max}\) and \(V_{min}\) are precisely what an oscilloscope displays when a trapezoidal or envelope pattern is examined, so the index can be measured directly from the screen without knowing the carrier or message amplitudes separately. Overmodulation shows itself immediately: the envelope would have to go negative for \(m\gt1\), so \(V_{min}\) flattens at zero and the envelope folds — visible distortion that a diode detector cannot undo.

Hence, m = (V_max − V_min)/(V_max + V_min).