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Question: What is the range of wavelengths of the electromagnetic waves that constitute visible radiation?...

What is the range of wavelengths of the electromagnetic waves that constitute visible radiation?

Explanation

Solution

The oscillations of the electric and the magnetic field vectors in directions perpendicular to each other is known as the electromagnetic wave. The features of the electromagnetic waves are:
1. They travel with the speed of light in vacuum.
2. It is a transverse wave.
3. The direction of the electric field vector is perpendicular to the propagation of wave and wave energy.
4. They do not require any medium to travel

The electromagnetic spectrum consists of a continuous band of different wavelength ranges which have different properties. Pertaining to the difference in the properties they have different uses in the real world.
The visible region of the spectrum lies between the ultraviolet region and the infrared region.

Complete step by step solution:
The electromagnetic spectrum is a continuous band of different wavelengths of light. Each band has different inherent properties. These bands along with their wavelengths are listed below:

Name of the bandWavelength λ\lambda (in m)
Gamma rays<10−12 < {10^{ - 12}}
X rays10−9−10−12{10^{ - 9}} - {10^{ - 12}}
Ultraviolet region4×10−7−10−94 \times {10^{ - 7}} - {10^{ - 9}}
Visible region7.5×10−7−4×10−77.5 \times {10^{ - 7}} - 4 \times {10^{ - 7}}
Infrared region2.5×10−6−2.5×10−52.5 \times {10^{ - 6}} - 2.5 \times {10^{ - 5}}
Microwaves2.5×10−5−10−32.5 \times {10^{ - 5}} - {10^{ - 3}}
Radio waves>10−3 > {10^{ - 3}}

Now if we go through the above table, we can deduce that the range of the electromagnetic waves that constitute the visible region is the wavelength band having extremum values as 7.5×10−7−4×10−77.5 \times {10^{ - 7}} - 4 \times {10^{ - 7}} .
So, in this range the electromagnetic waves are visible.

Note:
Since wavelength and frequency are inversely proportional to each other, their relations with the energy which the wave carries are also inverse. While the energy is directly proportional to the frequency, it is inversely proportional to the wavelength. So, the energy carried by the visible region is more than the energy carried by in the infrared region.