The intensity of gamma radiation from a given source is I. On passing through 36 mm of lead, it is reduced to 1/8. The thickness of lead which will reduce the intensity to 1/2 will be
As we know, the range of wavelength of visible radiation is from 400 nm to 700 nm. The radiation having wavelength greater than 700 nm is termed as infra-red radiations...
(a) point source
EXPLANATION:
Only for a point source the intensity can be written as I = E/4πr², where E is the strength of the source. Hence E ∝ 1/r².
The correct answer is (c) 9 : 4
EXPLANATION:
Since the resultant field at a point is given as
E₀² = E₁² + E₂² + 2E₁E₂ cosδ
Where E₁ and E₂ = amplitudes...
The correct answer is (a) (c)
(a) the maximum intensity of radiation will be near the frequency 2v0
(c) the total energy emitted will increase by a factor of 16.