Physics

Electromagnetic Waves and Spectrum

670 Questions

Electromagnetic waves and spectrum questions test a candidate's understanding of radiation frequencies, wavelengths, and the properties of different rays like infrared, ultraviolet, and visible light. Concepts also cover practical applications in astronomy and the fundamental speed of light calculations. This physics topic appears regularly in general science sections of major competitive examinations.

UV rays propertiesElectromagnetic radiation frequencyWavelength identificationSpeed of light calculationsBlack body radiation

Electromagnetic Waves and Spectrum Questions

Multiple choice

The distance between two successive peaks is called

  1. radiation length

  2. wavelength

  3. therapy length

  4. wave distance
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A wavelength is defined as the distance between two successive peaks of a wave. This is a fundamental concept in physics and light theory.

Multiple choice

What is a characteristic of short wavelengths?

  1. Low frequency

  2. Penetrate less

  3. Have less energy

  4. Fewer waves per length
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Short wavelengths have higher frequency and higher energy, but they have less penetrating power than longer wavelengths.

Multiple choice

The electromagnetic radiation with the lowest known frequency is:

  1. Gamma Rays

  2. Cosmic Rays

  3. Radio waves

  4. Microwaves
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The electromagnetic spectrum is ordered by frequency. Radio waves have the longest wavelengths and therefore the lowest frequencies, while gamma rays have the shortest wavelengths and highest frequencies.

Multiple choice

Transition from a high energy to a low energy state in an atom or molecule results in:

  1. Emission of a photon at a particular wavelength

  2. Absorption of a photon at a particular wavelength

  3. A red-shift

  4. A blue-shift
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When an electron drops from a higher energy level to a lower one, it loses energy. This energy is released as a photon with a frequency corresponding to the energy difference between the two states.

Multiple choice

How are Frequency and Wavelength related?

  1. As Wavelength Increases, Frequency Decreases

  2. As Wavelength Increases, Frequency Increases

  3. As Wavelength Decreases, Frequency Decreases

  4. They are NOT Related
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a constant wave speed, frequency and wavelength are inversely proportional (v = f * lambda). As wavelength increases, frequency must decrease.

Multiple choice

Which has Higher Frequency? UV-Rays or Microwaves?

  1. Ultraviolet rays

  2. Microwaves

  3. Electromagnetic Spectrum Rays

  4. None of the Above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Ultraviolet rays have a much higher frequency than microwaves, which are on the low-frequency end of the spectrum.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

Wavelength of light used in an optical instrument are $\lambda _1 = 4000 A^o  and \lambda _2 = 5000 A^0$, then ratio of their respective resolving powers (corresponding to $\lambda _1   \ and  \ \lambda _2$) is

  1. 16:25

  2. 9:1

  3. 4:5

  4. 5:4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Resolving power of an optical instrument $\displaystyle \propto \dfrac {1}{\lambda}$


$\displaystyle \dfrac{\text{Resolving  power  at} \lambda _1} {\text{Resolving  power  at} \lambda _2}=\dfrac {\lambda _1}{\lambda _2}$

$\displaystyle \left [\text{Limit of resolution}  \propto \dfrac{1}{\text{resolving  power}}\right]$

$\therefore$  Ratio  of  resolving  power $= \displaystyle \dfrac {5000}{4000} = \dfrac {5}{4} = 5 : 4$

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

Wavelength of light used in an optical instrument  are $\lambda _1 = 4000 \mathring { A } $ and $ \lambda _2 = 5000 \mathring { A} $ then ratio of their respective resolving powers(corresponding to $\lambda _1$ and $ \lambda _2$) is

  1. 16:25

  2. 9:1

  3. 4:5

  4. 5:4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know, 
$ Resolving\ Power\ \propto \dfrac{1}{\lambda} $

Given, 
$ \lambda _1 = 4000 \mathring{A} $
$ \lambda _2 = 5000 \mathring{A} $

$ \Rightarrow \dfrac{RP _{\lambda _1}}{RP _{\lambda _2}} = \dfrac{ \lambda _2}{ \lambda _1} $

$ \Rightarrow \dfrac{RP _{\lambda _1}}{RP _{\lambda _2}} = \dfrac{5000}{4000} = \dfrac{5}{4} $

$ \Rightarrow RP _{\lambda _1} : RP _{\lambda _2} = 5:4 $

Hence, the correct answer is OPTION D.