Step 1: Define modulation index in AM.
In amplitude modulation (AM), the amplitude of the carrier wave is varied in proportion to the message signal. The modulation index $ m $ (also called depth of modulation) is defined as: \[ m = \frac{A_m}{A_c} \] where $ A_m $ is the peak amplitude of the message signal and $ A_c $ is the peak amplitude of the carrier wave.
Step 2: Identify all given quantities.
Message signal: frequency $ f_m = 8\text{ kHz} $, peak voltage $ V_m = 12\text{ V} $. Carrier wave: frequency $ f_c = 1.2\text{ MHz} $, peak voltage $ V_c = 20\text{ V} $. Note that the frequencies are irrelevant for computing the modulation index - only the amplitudes matter.
Step 3: Substitute into the formula.
\[ m = \frac{V_m}{V_c} = \frac{12}{20} = 0.6 \]
Step 4: Interpret the result.
A modulation index of 0.6 (or 60%) means the carrier amplitude varies between $ A_c(1 - m) = 8\text{ V} $ and $ A_c(1 + m) = 32\text{ V} $. Since $ m < 1 $, this is an under-modulated signal with no distortion.
Step 5: Note the frequency condition.
For proper AM, the carrier frequency must be much greater than the message frequency: $ f_c \gg f_m $, i.e., $ 1.2\text{ MHz} \gg 8\text{ kHz} $. This condition is satisfied here.
Step 6: Final answer.
\[ \boxed{m = 0.6} \]