Assertion (A) : Light is electromagnetic wave in nature, these may be x-rays, radio waves, microwaves etc. Reason (R) : The amount of energy depends on the intensity of the light rays. The energy associated with each photon is proportional to the frequency. Select your answer using the codes given below :
Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Take the assertion first. Visible light is one narrow band of the electromagnetic spectrum, and x-rays, microwaves and radio waves are others. All of them are the same physical phenomenon — coupled oscillating electric and magnetic fields obeying Maxwell's equations and travelling at
\(c=\dfrac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}=3\times10^{8}\ \text{m/s}\)
differing only in frequency, so the assertion is true.
| Radiation | Frequency | Photon energy |
|---|---|---|
| Radio | kHz – GHz | Nano-eV to μeV |
| Microwave | 1 – 300 GHz | μeV to meV |
| Visible | 4 – 8 × 1014 Hz | 1.6 – 3.3 eV |
| X-ray | 1017 – 1019 Hz | keV — ionising |
The reason is true too. Total energy delivered per second does scale with intensity, since intensity counts photons per unit area per unit time; and the energy of an individual photon is
\(E=h\nu=\dfrac{hc}{\lambda}\)
These are the two independent statements of the photon picture, and they are exactly what the photoelectric effect demonstrates: raising the intensity releases more electrons but never more energetic ones, while raising the frequency raises the maximum kinetic energy through \(K_{max}=h\nu-\phi\).
The reason nevertheless does not explain the assertion. (A) is a claim about the wave nature of light and the unity of the spectrum; (R) is a claim about the particle nature and how energy is apportioned. Knowing that \(E=h\nu\) tells us nothing about why light is an electromagnetic wave — if anything it belongs to the complementary description. The two statements sit on opposite sides of wave-particle duality, which is precisely why the code is 2 rather than 1.
A caution on the assertion's wording. Saying that light "may be x-rays, radio waves, microwaves" is loose: light strictly means the visible band, and the correct statement is that all of these are electromagnetic radiation. The examiner's intent is clearly the broader sense, so the statement is taken as true.
Hence, both (A) and (R) are true, but (R) is not the correct explanation of (A).
If the axial ratio of an electromagnetic wave is zero dB. Then, the wave is:
A plane wave is propagating in a uniform medium with propagation constant \((0.6\pi + j\,0.4\pi)\) its wavelength is
Given below are two statements :
Statement I : Phase velocity of an electromagnetic wave in a bounded medium can be greater than the speed of light in an unbounded medium.
Statement II : Frequency of electromagnetic wave is affected by the permittivity of the medium.
In the light of the above statements, choose the correct answer from the options given below :
Assertion (A) : The ionosphere is that region of the earth’s atmosphere in which the constituent gases are ionised by radiations from outer space.
Reason (R) : Throughout the ionosphere, there are several layers in which ionization density either reaches maximum or remains almost constant. It depend on the intensity of sun. It also depends upon the atmospheric pressure.
The free space wave number ‘k0’ is defined as
Ultra High Frequency (UHF) spectrum is defined as :
Frequency in UHF range propagated by means of
For sky waves, following statements are given:
(A) n > 1, this shows 81 \(\rm\frac{N}{f^2}\) positive
(B) n > 1, show 81 \(\rm\frac{N}{f^2}\) Negative
(C) n < 1 shows 81 \(\rm\frac{N}{f^2}\) < 1
(D) v g x v p= c 2
(E) n = 0 shows 81 \(\rm\frac{N}{f^2}\) = 1, f = f c
Choose the correct answer from the options given below:
If the Polarization vector is given as N and the Direction of propagation is given as K then which one of the following relations is correct?
The wave length (λ) in meters of an electromagnetic wave is related to its frequency (f) in MHz as:
Bending of light wave as it passes between material of different optical density
The wave impedance of a medium is equal to: