Multiple-slit diffraction - class-XII

$\because n=3\ \therefore 2n\pi =2\times 3\times \pi \ \quad \quad \quad \quad \quad =6\pi $

The phase difference delta phi = (2 pi / lambda) * path difference. For path difference lambda/6, delta phi = pi/3 radians (60 degrees). Intensity I = I_max * cos^2(delta phi / 2), which gives I' = I_0 * cos^2(pi/6) = I_0 * (3/4), so the ratio I'/I_0 is 3:4.

For destructive interference (minima), the path difference must be an odd multiple of lambda/2, specifically (2n-1) * lambda/2. For the fifth order minimum (n=5), the path difference is (25 - 1) * lambda/2 = 9lambda/2.

The condition for dark fringes in YDSE is path difference = (2n-1) * lambda/2. For the third dark fringe (n=3), path difference = (23-1) * lambda/2 = 5lambda/2. Phase difference phi = (2pi/lambda) * path difference = (2pi/lambda) * (5lambda/2) = 5pi.

Candle power- It is the rating of light output at source. It is the the candle's luminous intensity in a particular direction.

The half period zone is provided by an optical device known as zone plate. It is simply a plane parallel gloss plate having concentric circles of radii accurately proportional to the square roots of the consecutive natural numbers 1,2,3 ... etc. The area is given by $\pi \gamma ^{2}$. Hence the
areas are $\pi , 2\pi , 3\pi , 4\pi$,---
The phase difference is thus $\pi$ between each successive half period zones of strips.

If there is no phase difference let
the Intensity equal to maximum Intensity be $I _{0}$
For a phase difference $\theta $ then
Intensity at a point equidistant from the two
slits is given by
$\dfrac{I _{0}}{2}(1+cos^{2}\theta )$
As $\theta =\pi /2$
we get $\dfrac{I _{0}}{2}(1+0)=\dfrac{I _{0}}{2}$
Thus Intensity $=\dfrac{I _{0}}{2}$

If the angle between two coherent sources is $\theta $ then the Intensity after Interference is given by $I _{1}+I _{2}+2\sqrt{I _{1}I _{2}} cos \theta $. For constructive Interference $\theta =2n\pi $ Thus Intensity is $I _{1}+I _{2}+2\sqrt{I _{1}I _{2}}$

Given $\cfrac { { I } _{ 1 } }{ { I } _{ 2 } } =\beta $

The angle between two light waves $={ \alpha  } _{ 1 }-{ \alpha  } _{ 2 }$

$I=I _1+I _2+2\sqrt{I _1I _2}\cos\theta =I _o+4I _o+2\sqrt{(I _o\times 4I _o)}\cos\pi =I _o$
Hence (A) is correct.

For minima , path difference=$\dfrac{(2n-1)\lambda}{2}$........(1)

Given that,

$ {{y} _{1}}=4\sin \omega t $

$ {{y} _{2}}=3\cos \omega t $

Amplitude of first and second wave is 4 cm and 3 cm. So, the amplitude of resultant wave is

$ {{y}^{'}}=\sqrt{{{(4)}^{2}}+{{(3)}^{2}}} $

$ =5\,cm $

For single slit diffraction, the condition for dark fringes is asin(theta) = nlambda. Given a = 24*10^-5 cm = 2400 nm, n=2, theta=30 degrees. 2400 * sin(30) = 2 * lambda => 2400 * 0.5 = 2 * lambda => 1200 = 2 * lambda => lambda = 600 nm = 6000 Angstrom.

$\begin{array}{l} { { { I } } _{ \max   } }={ \left( { \sqrt { { I _{ 1 } } } +\sqrt { { I _{ 2 } } }  } \right) ^{ 2 } } \ ={ I _{ 1 } }+{ I _{ 2 } }+2\sqrt { { I _{ 1 } }{ I _{ 2 } } }  \ { I _{ \min   } }={ I _{ 1 } }+{ I _{ 2 } }-2\sqrt { { I _{ 1 } }{ I _{ 2 } } }  \ \therefore \dfrac { { { { { I } } _{ \max   } }-{ { { I } } _{ \min   } } } }{ { { I _{ \max   } }+{ { { I } } _{ \min   } } } } =\dfrac { { 4\sqrt { { I _{ 1 } }{ I _{ 2 } } }  } }{ { 2\left( { { I _{ 1 } }+{ I _{ 2 } } } \right)  } }  \ =\dfrac { { 2\sqrt { { I _{ 1 } }{ I _{ 2 } } }  } }{ { \left( { { I _{ 1 } }+{ I _{ 2 } } } \right)  } }  \ { I _{ 1 } }=1 \ { I _{ 2 } }=\alpha  \ =\dfrac { { 2\sqrt { \alpha  }  } }{ { 1+\alpha  } }  \ \therefore , , Option, , \left( B \right) , , is, , correct, . \end{array}$

Destructive interference occurs when the difference is an odd multiple of $\pi$ ,

 $intensity \ \alpha \ (amplitude)^{2}$
so maximum intensity is proportional to $a^{2} _{1}+a^{2} _{2}$
option $C$ is correct 

As R$^{2}$ = a$^{2}$ + b$^{2}$ + 2 ab cos $\phi$
$\therefore$ I$ _{max}$ = I$ _1$ + I$ _2$ + 2$\sqrt{I _1I _2}$ cos 0$^{o}$ 
= I$ _1$ + I$ _2$ + 2$\sqrt{I _1I _2}$

In YDSE, intensity from both slits are same ,
So, $I _{0}=I _{max}=I+I+2\sqrt{II}=4I$
Now at path difference $\dfrac{\lambda }{4}$,
$\Delta \phi =\dfrac{2\pi }{4}=\dfrac{\pi }{2}$
So, $I=I+I+2\sqrt{\pm I}cos(\dfrac{\pi }{2})$
$=2I+2\sqrt{II}(0)    (\because cos\pi /2=0)$
$=2I$
So, $I=\dfrac{I _{0}}{2}$

phase difference corresponding to path difference of $\lambda /3$ is
$\phi =\dfrac{2\pi }{3}$
So, $I=k cos^{2}(\dfrac{2\pi /3}{2})     (\because I=I _{0} cos^{2}(\phi /2))$
         $=k(+1/2)^{2}     (\because cos \pi /3=1/2)$
         $=\dfrac{k}{4}$

Intensity at a point due to interference of beams of intensities $I _1,I _2$ with a phase difference $\phi$ between them=$I _1+I _2+2\sqrt{I _1I _2}cos\phi$

Maximum intensity,
     $I=4{ I } _{ 0 }\cos ^{ 2 }{ \left( \dfrac { \delta  }{ 2 }  \right)  } $
$\Rightarrow I\propto \cos ^{ 2 }{ \left( \dfrac { \delta  }{ 2 }  \right)  } $

The correct answer is option(C).

Resultant intensity at a point, $I _R= I _1+ I _2 +2\sqrt{I _1I _2} cos \phi$

For coherent sources, intensity at the center with equal amplitudes A is I1 = 4A^2. For incoherent sources, intensities simply add up, so I2 = A^2 + A^2 = 2A^2. The ratio I1 / I2 is therefore 4A^2 / 2A^2 = 2.

The resultant intensity is given by:
$I={I} _{1}+{I} _{2}+2\sqrt{{I} _{1}{I} _{2}}\cos{\phi}$
Thus difference is given by:
${I} _{A}-{I} _{B}=2\sqrt{{I} _{1}{I} _{2}}(\cos{\dfrac{\pi}{2}}-\cos{\pi})=2\sqrt{4{I}^{2}}\times1=4I$