Advertisements
Advertisements
प्रश्न
If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.
Advertisements
उत्तर
We know that intensity is directly proportional to the square of an amplitude
`I prop a^2`
if `I_1 = I/2`
if intensity reduced to 50%, the amplitude will be `a/sqrta2`
then `r = sqrt2`
`I_"max"/I_"min" = (r+1)^2/(r-1)^2 = (sqrt2 + 1)^2/(sqrt2 - 1)^2`
`I_"max"/I_"min" = ((2.414)/(0.414))^2 = (5.83)^2`
`I_"max"/I_"min"` = 33.98 ≅ 34
APPEARS IN
संबंधित प्रश्न
In young’s double slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence, deduce the expression for the fringe width.
Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.
Two polaroids ‘A’ and ‘B’ are kept in crossed position. How should a third polaroid ‘C’ be placed between them so that the intensity of polarized light transmitted by polaroid B reduces to 1/8th of the intensity of unpolarized light incident on A?
Can we perform Young's double slit experiment with sound waves? To get a reasonable "fringe pattern", what should be the order of separation between the slits? How can the bright fringes and the dark fringes be detected in this case?
If Young's double slit experiment is performed in water, _________________ .
White coherent light (400 nm-700 nm) is sent through the slits of a Young's double slit experiment (see the following figure). The separation between the slits is 0⋅5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1⋅0 mm away (along the width of the fringes) from the central line. (a) Which wavelength(s) will be absent in the light coming from the hole? (b) Which wavelength(s) will have a strong intensity?
In a Young's double slit experiment, \[\lambda = 500\text{ nm, d = 1.0 mm and D = 1.0 m.}\] Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.
"If the slits in Young's double slit experiment are identical, then intensity at any point on the screen may vary between zero and four times to the intensity due to single slit".
Justify the above statement through a relevant mathematical expression.
A slit of width 0.6 mm is illuminated by a beam of light consisting of two wavelengths 600 nm and 480 nm. The diffraction pattern is observed on a screen 1.0 m from the slit. Find:
- The distance of the second bright fringe from the central maximum pertaining to the light of 600 nm.
- The least distance from the central maximum at which bright fringes due to both wavelengths coincide.
A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is 'x' nm. The value of 'x' to the nearest integer is ______.
