Respuesta :
Answer:
Explanation:
Given
Space between two slits [tex]d=0.260 mm[/tex]
Distance between slit and screen is [tex]D=0.7 m[/tex]
Wavelength [tex]\lambda =660 nm[/tex]
Distance of n th drak fringe From central Fringe is given by
[tex]y_n=\frac{(2n-1)D\lambda }{2d}[/tex]
For first minimum i.e. n=1
[tex]y_1=\frac{1\times 0.7\times 660\times 10^{-9}}{2\times 0.260\times 10^{-3}}[/tex]
[tex]y_1=888.46\times 10^{-6}[/tex]
[tex]y_1=0.88 mm[/tex]
(b)Distance where intensity reduces to [tex]\frac{I_0}{2}[/tex]
Intensity is given by
[tex]I=I_o\cos ^{2}\frac{\phi }{2}[/tex]
where [tex]\phi [/tex]phase difference
For [tex]I=\frac{I_0}{2}[/tex]
[tex]\frac{1}{2}=\cos ^{2}\frac{\phi }{2}[/tex]
[tex]\cos \frac{\phi }{2}=\frac{1}{\sqrt{2}}[/tex]
thus [tex]\phi =\frac{\pi }{2}[/tex]
For Phase difference of [tex]\frac{\pi }{2}[/tex] Path difference is
[tex]x=\frac{\lambda }{2\pi}\times \phi[/tex]
[tex]x=\frac{\lambda }{2\pi }\times \frac{\pi }{2}[/tex]
[tex]x=\frac{\lambda }{4}[/tex]
Now its Position From Central Point is given by
[tex]y=\frac{D}{d}\cdot x[/tex]
[tex]y=\frac{0.7}{0.260\times 10^{-3}}\cdot \frac{660\times 10^{-9}}{4}[/tex]
[tex]y=444.23\times 10^{-6} m[/tex]
[tex]y=0.44 mm[/tex]
Young's experiment is used to find the distance of n'th number dark fringe from the central fringe of screen.
- (a)The distance on the screen from the center of the central maximum to the first minimum is 0.88 mm.
- (b) The distance on the screen from the center of the central maximum to the point where the intensity has fallen to [tex]\dfrac{I_0}{2}[/tex] is 0.44 mm.
What is young's experiment?
Young's experiment is used to find the distance of n'th number dark fringe from the central fringe of screen.
It can be find out with the following formula given as,
[tex]y_n=\dfrac{(2n-1)D\lambda}{2d}[/tex]
Here, [tex]\lambda[/tex] is the wavelength of the light and [tex]D[/tex] is the distance of the screen to the slits.
Given information-
The distance between the two slits is 0.260 mm.
The distance between the slits and screen is 0.700 m.
The wavelength of the coherent light is 660 nm.
The intensity at the center of the central maximum ([tex]\theta=0^o[/tex]) is [tex]I_0[/tex]
- (a)The distance on the screen from the center of the central maximum to the first minimum-
Use the above formula for the first minimum as,
[tex]y_1=\dfrac{(2(1)-1)0.7\times660\times10^{-9}}{2\times0.26\times10^{-3}}\\y_1=0.88\rm mm[/tex]
Thus, the distance on the screen from the center of the central maximum to the first minimum is 0.88 mm.
- (b) The distance on the screen from the center of the central maximum to the point where the intensity has fallen to [tex]\dfrac{I_0}{2}[/tex]-
As the intensity has fallen to the [tex]I_0[/tex]/2, thus the value of path difference will become equal to the [tex]\pi/2[/tex].
The above formula for this path difference can be written as,
[tex]y=\dfrac{(2\times1)D}{2d}\times\dfrac{\lambda}{2\pi}\times\dfrac{\pi}{2}\\y=\dfrac{(0.7}{0.260\times10^{-3}}\times\dfrac{660\times10^{-9}}{2\pi}\times\dfrac{\pi}{2}\\y=0.44\rm nm[/tex]
Thus, the distance on the screen from the center of the central maximum 0.44 mm.
Hence,
- (a)The distance on the screen from the center of the central maximum to the first minimum is 0.88 mm.
- (b) The distance on the screen from the center of the central maximum to the point where the intensity has fallen to [tex]\dfrac{I_0}{2}[/tex] is 0.44 mm.
Learn more about the young's experiment here;
https://brainly.com/question/4449144
