Q. 264.7( 4 Votes )

<span lang="EN-US

Answer :

a) In Set A, as the phase difference between the two waves emanating from the slits does not change with time so alternate bright and dark bands would be seen on the screen as per Young’s double slit experiment as two slits are acting as coherent sources whereas in Set B, the Phase difference changes rapidly with time, the light waves coming out from two independent sources of light will not have any fixed phase relationship and would be incoherent, when this happens, the intensities on the screen will add up and only bright light will be seen.


b) Given: -


The waves emanating from the two slits have the same amplitude A and same wavelength λ


Derivation: -


Let the displacement produced by wave emanating from slit 1 be y1 and from slit 2 be y2


So, we can write,


y1=Acosωt


And,


y2=Acos(ωt+ϕ)


Where is the phase difference between the two sources, and A is the amplitude,


the resultant displacement will be given by,


y = y1 + y2


so, after substituting the values we get,


using the trigonometric identity,



We get,



So, the amplitude of the resultant displacement is,



As we know that the intensity is directly proportional to the square of the amplitude, so,


I0A2


And


IA'2


So, we can write,



Which gives,



Conclusions: -


The resultant intensity is given as,



For set A,


As is constant, the value of I remains constant for a particular band on the screen, which leads to formation of alternate bright and dark bands.


For set B,


As is changing rapidly with time, the value of I changes rapidly with time on the screen. The value of I changes so rapidly from 0 to 4I0 and again back to zero so rapidly, so it seems to the human eye that the intensity is constant as I = 4I0 and it seems that constructive inference is occurring.


OR


a) Given: -


The intensity of the incident light I0


Derivation: -


After passing the first polaroid (P1) the intensity be I1


We know that I1 reduces to half


38.JPG


After passing the third polaroid (P3) the intensity be I2


We can write by malus’ law as,



Also the angle between the pass axis of P3 and P2 is 90 - ϴ,


So the intensity of light transmitted through P2 is,






a) When unpolarized light is incident on a plane glass surface, the reflected light is polarized with its electric vector perpendicular to the plane of incidence when the refracted and reflected rays make a right angle with each other. Thus, when reflected wave is perpendicular to the refracted wave, the reflected wave is a totally polarized wave. The angle of incidence in this case is called Brewster’s angle and is denoted by iB. We can see that iB is related to the refractive index of the denser medium.


39.JPG


Since we have,



from Snell’s law we get,




So, finally,


μ=taniB


The above expression is called the Brewster’s Law.


The state of the above light is partially polarized because When such light is viewed through a rotating analyzer, one sees a maximum and a minimum of intensity but not complete darkness.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Sketch the graphsPhysics - Board Papers

Define the term wPhysics - Board Papers

When unpolarised Physics - Board Papers

When a potential Physics - Board Papers

<span lang="EN-USPhysics - Exemplar

<span lang="EN-USPhysics - Exemplar