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Electromagnetic Waves: JEE Main Physics Question with Solution For an amplitude modulated wave the minimum amplitude is
3 V 3\text{ V} 3 V , while the modulation index is
60 % 60\% 60% . The maximum amplitude of the modulated wave is:
Step-by-step solution View Correct answer
Using the relation between modulation index and extreme amplitudes, the maximum amplitude is 12 V. Option analysis
Why each option works or fails A · 15 V 15\text{V} 15 V Believing that the ratio ( A max + A min ) / ( A max − A min ) (A_{\text{max}} + A_{\text{min}})/(A_{\text{max}} - A_{\text{min}}) ( A max + A min ) / ( A max − A min ) μ \mu μ 15 V 15\text{ V} 15 V μ = ( A max − A min ) / ( A max + A min ) \mu = (A_{\text{max}} - A_{\text{min}})/(A_{\text{max}} + A_{\text{min}}) μ = ( A max − A min ) / ( A max + A min )
B · 12 V 12\text{V} 12 V None. This is the correct calculation: μ = A max − A min A max + A min ⟹ 0.6 = A max − 3 A max + 3 ⟹ A max = 12 V \mu = \frac{A_{\text{max}} - A_{\text{min}}}{A_{\text{max}} + A_{\text{min}}} \implies 0.6 = \frac{A_{\text{max}} - 3}{A_{\text{max}} + 3} \implies A_{\text{max}} = 12\text{ V} μ = A max + A min A max − A min ⟹ 0.6 = A max + 3 A max − 3 ⟹ A max = 12 V
C · 10 V 10\text{V} 10 V Making an arithmetic error when clearing denominators, such as writing 0.6 A max + 3 = A max − 3 0.6 A_{\text{max}} + 3 = A_{\text{max}} - 3 0.6 A max + 3 = A max − 3 3 3 3 0.6 0.6 0.6 μ = 0.6 \mu = 0.6 μ = 0.6 0.6 ( A max + 3 ) = 0.6 A max + 1.8 0.6(A_{\text{max}} + 3) = 0.6A_{\text{max}} + 1.8 0.6 ( A max + 3 ) = 0.6 A max + 1.8
D · 5 V 5\text{V} 5 V Mistaking the minimum amplitude for the carrier amplitude A c A_c A c A max = A c ( 1 + μ ) ≈ 3 × 1.6 = 4.8 ≈ 5 V A_{\text{max}} = A_c(1 + \mu) \approx 3 \times 1.6 = 4.8 \approx 5\text{ V} A max = A c ( 1 + μ ) ≈ 3 × 1.6 = 4.8 ≈ 5 V A min A_{\text{min}} A min A min = A c − A m = A c ( 1 − μ ) A_{\text{min}} = A_c - A_m = A_c(1 - \mu) A min = A c − A m = A c ( 1 − μ ) A c A_c A c
Step Working
01 given A min = 3 V A_{\min} = 3\text{ V} A m i n = 3 V μ = 60 % = 0.6 \mu = 60\% = 0.6 μ = 60% = 0.6
02 find Maximum amplitude of the modulated wave, A max A_{\max} A m a x
03 visualise An AM wave's envelope varies between A max = A c + A m A_{\max} = A_c + A_m A m a x = A c + A m A min = A c − A m A_{\min} = A_c - A_m A m i n = A c − A m μ = A m / A c \mu = A_m / A_c μ = A m / A c
04 strategise Express A min A_{\min} A m i n A c A_c A c A m = 0.6 A c A_m = 0.6 A_c A m = 0.6 A c A c A_c A c A m A_m A m A max A_{\max} A m a x
05 execute A min = A c − 0.6 A c = 0.4 A c = 3 V ⟹ A c = 7.5 V A_{\min} = A_c - 0.6 A_c = 0.4 A_c = 3\text{ V} \implies A_c = 7.5\text{ V} A m i n = A c − 0.6 A c = 0.4 A c = 3 V ⟹ A c = 7.5 V A m = 0.6 × 7.5 = 4.5 V A_m = 0.6 \times 7.5 = 4.5\text{ V} A m = 0.6 × 7.5 = 4.5 V A max = 7.5 + 4.5 = 12 V A_{\max} = 7.5 + 4.5 = 12\text{ V} A m a x = 7.5 + 4.5 = 12 V
✓ verify Check: μ = ( A max − A min ) / ( A max + A min ) = ( 12 − 3 ) / ( 12 + 3 ) = 9 / 15 = 0.6 = 60 % \mu = (A_{\max} - A_{\min}) / (A_{\max} + A_{\min}) = (12 - 3) / (12 + 3) = 9 / 15 = 0.6 = 60\% μ = ( A m a x − A m i n ) / ( A m a x + A m i n ) = ( 12 − 3 ) / ( 12 + 3 ) = 9/15 = 0.6 = 60%
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Exam relevance JEE Main · Physics
Concepts assessed Physics
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Editorial review 9 September 2026 Quick checks
Students also ask Why isn't modulation index simply the ratio A min / A max A_{\min} / A_{\max} A m i n / A m a x By definition, the modulation index is μ = A m / A c = ( A max − A min ) / ( A max + A min ) \mu = A_m / A_c = (A_{\max} - A_{\min}) / (A_{\max} + A_{\min}) μ = A m / A c = ( A m a x − A m i n ) / ( A m a x + A m i n )
Answer Using the relation between modulation index and extreme amplitudes, the maximum amplitude is 12 V.
Why each option works or fails A: 15 V 15\text{V} 15 V - Believing that the ratio ( A max + A min ) / ( A max − A min ) (A_{\text{max}} + A_{\text{min}})/(A_{\text{max}} - A_{\text{min}}) ( A max + A min ) / ( A max − A min ) μ \mu μ 15 V 15\text{ V} 15 V μ = ( A max − A min ) / ( A max + A min ) \mu = (A_{\text{max}} - A_{\text{min}})/(A_{\text{max}} + A_{\text{min}}) μ = ( A max − A min ) / ( A max + A min ) B · correct: 12 V 12\text{V} 12 V - None. This is the correct calculation: μ = A max − A min A max + A min ⟹ 0.6 = A max − 3 A max + 3 ⟹ A max = 12 V \mu = \frac{A_{\text{max}} - A_{\text{min}}}{A_{\text{max}} + A_{\text{min}}} \implies 0.6 = \frac{A_{\text{max}} - 3}{A_{\text{max}} + 3} \implies A_{\text{max}} = 12\text{ V} μ = A max + A min A max − A min ⟹ 0.6 = A max + 3 A max − 3 ⟹ A max = 12 V C: 10 V 10\text{V} 10 V - Making an arithmetic error when clearing denominators, such as writing 0.6 A max + 3 = A max − 3 0.6 A_{\text{max}} + 3 = A_{\text{max}} - 3 0.6 A max + 3 = A max − 3 3 3 3 0.6 0.6 0.6 μ = 0.6 \mu = 0.6 μ = 0.6 0.6 ( A max + 3 ) = 0.6 A max + 1.8 0.6(A_{\text{max}} + 3) = 0.6A_{\text{max}} + 1.8 0.6 ( A max + 3 ) = 0.6 A max + 1.8 D: 5 V 5\text{V} 5 V - Mistaking the minimum amplitude for the carrier amplitude A c A_c A c A max = A c ( 1 + μ ) ≈ 3 × 1.6 = 4.8 ≈ 5 V A_{\text{max}} = A_c(1 + \mu) \approx 3 \times 1.6 = 4.8 \approx 5\text{ V} A max = A c ( 1 + μ ) ≈ 3 × 1.6 = 4.8 ≈ 5 V A min A_{\text{min}} A min A min = A c − A m = A c ( 1 − μ ) A_{\text{min}} = A_c - A_m = A_c(1 - \mu) A min = A c − A m = A c ( 1 − μ ) A c A_c A c Step-by-step solution given: A min = 3 V A_{\min} = 3\text{ V} A m i n = 3 V μ = 60 % = 0.6 \mu = 60\% = 0.6 μ = 60% = 0.6 find: Maximum amplitude of the modulated wave, A max A_{\max} A m a x visualise: An AM wave's envelope varies between A max = A c + A m A_{\max} = A_c + A_m A m a x = A c + A m A min = A c − A m A_{\min} = A_c - A_m A m i n = A c − A m μ = A m / A c \mu = A_m / A_c μ = A m / A c strategise: Express A min A_{\min} A m i n A c A_c A c A m = 0.6 A c A_m = 0.6 A_c A m = 0.6 A c A c A_c A c A m A_m A m A max A_{\max} A m a x execute: A min = A c − 0.6 A c = 0.4 A c = 3 V ⟹ A c = 7.5 V A_{\min} = A_c - 0.6 A_c = 0.4 A_c = 3\text{ V} \implies A_c = 7.5\text{ V} A m i n = A c − 0.6 A c = 0.4 A c = 3 V ⟹ A c = 7.5 V A m = 0.6 × 7.5 = 4.5 V A_m = 0.6 \times 7.5 = 4.5\text{ V} A m = 0.6 × 7.5 = 4.5 V A max = 7.5 + 4.5 = 12 V A_{\max} = 7.5 + 4.5 = 12\text{ V} A m a x = 7.5 + 4.5 = 12 V verify: Check: μ = ( A max − A min ) / ( A max + A min ) = ( 12 − 3 ) / ( 12 + 3 ) = 9 / 15 = 0.6 = 60 % \mu = (A_{\max} - A_{\min}) / (A_{\max} + A_{\min}) = (12 - 3) / (12 + 3) = 9 / 15 = 0.6 = 60\% μ = ( A m a x − A m i n ) / ( A m a x + A m i n ) = ( 12 − 3 ) / ( 12 + 3 ) = 9/15 = 0.6 = 60% Shortcut: When to use it: Fast direct relation between A_max, A_min, and mu.
given: A min = 3 V A_{\min} = 3\text{ V} A m i n = 3 V μ = 0.6 \mu = 0.6 μ = 0.6
find: A max A_{\max} A m a x
strategise: Use the direct relation A max A min = 1 + μ 1 − μ \frac{A_{\max}}{A_{\min}} = \frac{1 + \mu}{1 - \mu} A m i n A m a x = 1 − μ 1 + μ
execute: A max = A min ( 1 + 0.6 1 − 0.6 ) = 3 × 1.6 0.4 = 3 × 4 = 12 V A_{\max} = A_{\min} \left(\frac{1 + 0.6}{1 - 0.6}\right) = 3 \times \frac{1.6}{0.4} = 3 \times 4 = 12\text{ V} A m a x = A m i n ( 1 − 0.6 1 + 0.6 ) = 3 × 0.4 1.6 = 3 × 4 = 12 V
verify: A max = 12 V > A min = 3 V A_{\max} = 12\text{ V} > A_{\min} = 3\text{ V} A m a x = 12 V > A m i n = 3 V ( 1 + 0.6 ) / ( 1 − 0.6 ) = 4 (1+0.6)/(1-0.6) = 4 ( 1 + 0.6 ) / ( 1 − 0.6 ) = 4