Answer
Verified
398.7k+ views
Hint: Intensity of a light beam depends on the amplitude of the light. When two coherent light beams merge, it can have different types of superpositions. The maximum intensity occurs when they are in-phase and minimum superposition occurs when they are completely out of phase.
Complete step-by-step answer:
Superposition of two light beams occurs when two light beams converge at a position. The resultant light beam has an amplitude which depends on the amplitude of the constituent waves and the phase of those waves.
The intensity of light waves depends on the amplitude of the wave in this manner.
$I\propto {{A}^{2}}$
When two light waves converge, we cannot add the intensity of the light waves.
We need to find the amplitude of the waves and then calculate the resultant amplitude of the new superimposed wave.
Let’s assume the amplitude of the wave that has intensity I, is A.
As intensity depends on the square of the amplitude, we can write
$I=k{{A}^{2}}$
Hence, for the other light wave amplitude can be given by,
$4I=k{{(2A)}^{2}}$
If the two light waves converge in-phase, the amplitudes can be added.
Hence, the resultant amplitude will be = (A+2A) = 3A.
So, the intensity will be,
$I'=k{{(3A)}^{2}}=9k{{A}^{2}}=9I$
When the two light waves converge out-of-phase, the amplitude will be,
(2A-A) =A
Hence, the intensity of the resultant light will be,
$I''=k{{(A)}^{2}}=k{{A}^{2}}=I$
So, the maximum intensity will be 9I and the minimum intensity will be I.
The correct answer is (C).
Note: There is another formula to calculate the maximum and minimum intensity of the light wave.
The maximum intensity can be given by,
$I={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}$
The minimum intensity can be given by,
$I={{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}$
The intensity due to a superposition can be given by,
$I={{({{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \theta )}^{2}}$
Where.
${{I}_{1}}$ is the intensity of one light beam
${{I}_{2}}$ is the intensity of the second light beam
$\theta $ phase difference between the two light beams
Complete step-by-step answer:
Superposition of two light beams occurs when two light beams converge at a position. The resultant light beam has an amplitude which depends on the amplitude of the constituent waves and the phase of those waves.
The intensity of light waves depends on the amplitude of the wave in this manner.
$I\propto {{A}^{2}}$
When two light waves converge, we cannot add the intensity of the light waves.
We need to find the amplitude of the waves and then calculate the resultant amplitude of the new superimposed wave.
Let’s assume the amplitude of the wave that has intensity I, is A.
As intensity depends on the square of the amplitude, we can write
$I=k{{A}^{2}}$
Hence, for the other light wave amplitude can be given by,
$4I=k{{(2A)}^{2}}$
If the two light waves converge in-phase, the amplitudes can be added.
Hence, the resultant amplitude will be = (A+2A) = 3A.
So, the intensity will be,
$I'=k{{(3A)}^{2}}=9k{{A}^{2}}=9I$
When the two light waves converge out-of-phase, the amplitude will be,
(2A-A) =A
Hence, the intensity of the resultant light will be,
$I''=k{{(A)}^{2}}=k{{A}^{2}}=I$
So, the maximum intensity will be 9I and the minimum intensity will be I.
The correct answer is (C).
Note: There is another formula to calculate the maximum and minimum intensity of the light wave.
The maximum intensity can be given by,
$I={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}$
The minimum intensity can be given by,
$I={{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}$
The intensity due to a superposition can be given by,
$I={{({{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \theta )}^{2}}$
Where.
${{I}_{1}}$ is the intensity of one light beam
${{I}_{2}}$ is the intensity of the second light beam
$\theta $ phase difference between the two light beams
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE