Advertisements
Advertisements
प्रश्न
The intensity at the central maxima in Young’s double slit experiment is I0. Find out the intensity at a point where the path difference is` lambda/6,lambda/4 and lambda/3.`
Advertisements
उत्तर
The intensity of central maxima is I0. Let I1 and I2 be the intensity emitted by the two slits S1 and S2, respectively.
The expression for resultant intensity is
`I=I_1+I_2+2sqrt(I_1I_2)cosphi`
For central maxima, I = I0 and Φ = 0
We assume I1 = I2
∴ I0=2I1+2I1 cos0=4I1
∴ I1 = I2= `I_0/4`
Now, when the path difference is `lambda/6`we get
`phi=(2pi)/lambdaxxp.d=(2pi)/lambdaxxlambda/6=pi/3`
`:.I'=I_1+I_2+2sqrt(I_1I_2)cos`
`:.I'=2I_0/4+2I_0/4xx1/2`
`:.I'=I_0/2+I_0/4=(3I_0)/4`
Similarly, when the path difference is `lambda/4`we get
`phi=(2pi)/lambdaxxp.d=(2pi)/lambdaxxlambda/4=pi/2`
`:.I'=I_1+I_2+2sqrt(I_1I_2)cos""pi/2`
`:.I'=2I_0/4+0`
`:.I'=I_0/2`
Finally, when the path difference is `lambda/3`we get
`phi=(2pi)/lambdaxxp.d=(2pi)/lambdaxxlambda/3=(2pi)/3`
`:.I'=I_1+I_2+2sqrt(I_1I_2)cos ""(2pi)/3`
`:.I'=2I_0/4+2I_0/4xx(-1/2)`
`:.I'=I_0/2-I_0/4=I_0/4`
APPEARS IN
संबंधित प्रश्न
A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young's double slit experiment on a screen placed 1 · 4 m away. If the two slits are separated by 0·28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.
A source emitting light of wavelengths 480 nm and 600 nm is used in a double-slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.
A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Slow that if the incident beam makes an angle \[\theta = \sin^{- 1} \left( \frac{\lambda}{2d} \right)\] with the normal to the plane of the slits, there will be a dark fringe at the centre P0 of the pattern.
Wavefront is ______.
In Young's double slit experiment shown in figure S1 and S2 are coherent sources and S is the screen having a hole at a point 1.0 mm away from the central line. White light (400 to 700 nm) is sent through the slits. Which wavelength passing through the hole has strong intensity?
Why is the diffraction of sound waves more evident in daily experience than that of light wave?
The central fringe of the interference pattern produced by the light of wavelength 6000 Å is found to shift to the position of the fourth bright fringe after a glass plate of refractive index 1.5 is introduced in the path of one of the beams. The thickness of the glass plate would be ______.
Two beams of light having intensities I and 41 interfere to produce a fringe pattern on a screen. The phase difference between the two beams are π/2 and π/3 at points A and B respectively. The difference between the resultant intensities at the two points is xl. The value of x will be ______.
In an interference experiment, a third bright fringe is obtained at a point on the screen with a light of 700 nm. What should be the wavelength of the light source in order to obtain the fifth bright fringe at the same point?
What does \(d\) denote in Young's double-slit experiment?
