JEE Main 2020 — Physics Electromagnetism

2020mcqeasy

Choose the correct option relating wavelengths of different parts of electromagnetic wave spectrum:

Official previous-year question

Held on 4 Sept 2020 · Verified 6 Jul 2026.

Options

A
${\lambda }_{\text{visible}}<{\lambda }_{\text{micro} \text{waves}}<{\lambda }_{\text{radio} \text{waves}}<{\lambda }_{\text{x-rays}}$
B
${\lambda }_{\text{radio} \text{waves}}>{\lambda }_{\text{micro} \text{waves}}>{\lambda }_{\text{visible}}>{\lambda }_{\text{x-rays}}$
C
${\lambda }_{\text{x-rays}}<{\lambda }_{\text{micro} \text{waves}}<{\lambda }_{\text{radio} \text{waves}}<{\lambda }_{\text{visible}}$
D
${\lambda }_{\text{visible}}>{\lambda }_{\text{x-rays}}>{\lambda }_{\text{radio} \text{waves}}>{\lambda }_{\text{micro} \text{waves}}$

Did you get this right?

Solution

First, recall the spectrum of electromagnetic waves, and we know that in the case of electromagnetic waves the frequency is inversely proportional to wavelength.

JEE Main 2020 Physics, Electromagnetic Waves — solution figure

Now from the above spectrum, we can say that, the relations of wavelength,

${\lambda }_{\text{radio waves}}>{\lambda }_{\text{micro waves}}>{\lambda }_{\text{visible}}>{\lambda }_{\text{x-rays}}$ .

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Options

${\lambda }_{\text{visible}}<{\lambda }_{\text{micro} \text{waves}}<{\lambda }_{\text{radio} \text{waves}}<{\lambda }_{\text{x-rays}}$

${\lambda }_{\text{radio} \text{waves}}>{\lambda }_{\text{micro} \text{waves}}>{\lambda }_{\text{visible}}>{\lambda }_{\text{x-rays}}$

${\lambda }_{\text{x-rays}}<{\lambda }_{\text{micro} \text{waves}}<{\lambda }_{\text{radio} \text{waves}}<{\lambda }_{\text{visible}}$

${\lambda }_{\text{visible}}>{\lambda }_{\text{x-rays}}>{\lambda }_{\text{radio} \text{waves}}>{\lambda }_{\text{micro} \text{waves}}$

Solution

First, recall the spectrum of electromagnetic waves, and we know that in the case of electromagnetic waves the frequency is inversely proportional to wavelength.

JEE Main 2020 Physics, Electromagnetic Waves — solution figure

Now from the above spectrum, we can say that, the relations of wavelength,

${\lambda }_{\text{radio waves}}>{\lambda }_{\text{micro waves}}>{\lambda }_{\text{visible}}>{\lambda }_{\text{x-rays}}$ .

JEE Main 2020 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism

JEE Main 2020 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism