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  • Visible height of wavelength cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction min. i

Visible height of wavelength 6000\times10^{-8} cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction min. is at 60 from the central max. If the first minimum is produced at \theta_{1}, the \theta_{1} is close to,
 
Option: 1 45^{o}
Option: 2 30^{o}
Option: 3 25^{o}  
Option: 4 20^{o}

Answers (1)

best_answer

 

 

Fraunhofer diffraction by a single slit -

Fraunhofer diffraction by a single slit

let's assume  a plane wave front is incident on a slit AB (of width b). Each and every part of the exposed part of the plane wave front (i.e. every part of the slit) acts as a source of secondary wavelets spreading in all directions. The diffraction is obtained on a screen placed at a large distance. (In practice, this condition is achieved by placing the screen at the focal plane of a converging lens placed just after the slit).

  •  The diffraction pattern consists of a central bright fringe (central maxima) surrounded by dark and bright lines (called secondary minima and maxima).
  •  At point O on the screen, the central maxima is obtained. The wavelets originating from points A and B meets in the same phase at this point, hence at O, intensity is maximum

 Secondary minima : For obtaining nth secondary minima at P on the screen, path difference between the diffracted waves 

                    \Delta x =b\sin{\theta }=n \lambda

 

  1. Angular position of nth secondary minima: 

              \sin{\theta } \approx \theta = \frac{n\lambda} {b}

       2. Distance of nth secondary minima from central maxima:

 x_{n}=D\cdot \theta =\frac{n\lambda D}{b}     where D = Distance between slit and screen. 

                                             f\approx D = Focal length of converging lens. 

Secondary maxima : For nth secondary maxima at P on the screen.

Path difference    \Delta x =b\sin{\theta }=(2n+1) \frac{\lambda}{2}  ; where n = 1, 2, 3 ..... 

(i) Angular position of nth secondary maxima 

\sin{\theta } \approx \theta \approx \frac{(2n+1)\lambda} {2b}

(ii) Distance of nth secondary maxima from central maxima: 

x_{n}=D\cdot \theta =\frac{(2n+1)\lambda D}{2b}

 Central maxima : The central maxima lies between the first minima on both sides.

(i) The Angular width d central maxima    =  2 \theta=\frac{2 \lambda}{b}
(ii) Linear width of central maxima    =2 x=2 D \theta=2 f \theta=\frac{2 \lambda f}{b}

 Intensity distribution: if the intensity of the central maxima is  $I_{0}$  then the intensity of the first and secondary maxima are
found to be   \frac{I_{0}}{22}$ and $\frac{I_{0}}{61}. Thus diffraction fringes are of unequal width and unequal intenstities.

(i) The mathematical expression for in intensity distribution on the screen is given by: 

I=I_{o}\left(\frac{\sin \alpha}{\alpha}\right)^{2}  where \alpha  is just a convenient connection between the angle \theta that locates a point on the viewing screening and light intensity I. 

  

$\phi=$  Phase difference between the top and bottom ray from the slit width b.
Also   \alpha=\frac{1}{2} \phi=\frac{\pi b}{\lambda} \sin \theta. 

(ii) As the slit width increases relative to wavelength the width of the control diffraction maxima decreases; that is, the light undergoes less flaring by the slit. The secondary maxima also decreases in width and becomes weaker.

(iii) If  b>>λ, the secondary maxima due to the slit disappear; we then no longer have single slit diffraction.

-

 

 

 

d\ sin\theta=path\ difference=2\lambda

So,

\\d\ sin\60=2\lambda\\\Rightarrow \frac{\lambda}{d}=\frac{\sqrt{3}}{4}

For first minima,

\\d\ sin\theta_2=\lambda\\sin\theta_2=\frac{\lambda}{d}=\frac{\sqrt{3}}{4}\\\Rightarrow \theta_2=sin^{-1}(\frac{\sqrt{3}}{4})=25.64^\circ=25^\circ

 

So option (3) is correct.

Posted by

Ritika Jonwal

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