Advertisements
Advertisements
प्रश्न
In a double-slit experiment using the light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.
Advertisements
उत्तर १
The angular width (θ) of the fringe in the double-slit experiment is given by,
`theta=lambda/"d"`
Where
d = Spacing between the slits
Given:
The wavelength of light, λ = 600 nm
The angular width of the fringe,
θ = 0.1° = `pi/1800` = 0.0018 rad
∴ d = `lambda/theta`
d = `(600xx10^(-9))/(18xx10^(-4))`
d = 0.33 × 10−3 m
उत्तर २
Wavelength of light used, λ = 6000 nm = 600 × 10−9 m
Angular width of fringe, θ = 0.1° =` 0.1 xx pi/180 = 3.14/1800 "rad"`
The angular width of a fringe is related to slit spacing (d) as:
θ = `lambda/"d"`
`"d"= lambda/θ`
= `(600 xx 10^(-9))/(3.14 /1800)`
= 3.44 × 10−4 m
Therefore, the spacing between the slits is 3.44 × 10−4 m.
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.
The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9 : 25. Find the ratio of the widths of the two slits.
In Young’s experiment, the ratio of intensity at the maxima and minima in an interference
pattern is 36 : 9. What will be the ratio of the intensities of two interfering waves?
Find the intensity at a point on a screen in Young's double slit experiment where the interfering waves have a path difference of (i) λ/6, and (ii) λ/2.
In Young’s experiment interference bands were produced on a screen placed at 150 cm from two slits, 0.15 mm apart and illuminated by the light of wavelength 6500 Å. Calculate the fringe width.
A Young's double slit experiment is performed with white light.
(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(c) The fringe next to the central will be red.
(d) The fringe next to the central will be violet.
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.
In Young’s double-slit experiment, show that:
`beta = (lambda "D")/"d"` where the terms have their usual meaning.
In Young's double-slit experiment, the two slits are separated by a distance of 1.5 mm, and the screen is placed 1 m away from the plane of the slits. A beam of light consisting of two wavelengths of 650 nm and 520 nm is used to obtain interference fringes.
Find the distance of the third bright fringe for λ = 520 nm on the screen from the central maximum.
In Young’s double slit experiment, what is the effect on fringe pattern if the slits are brought closer to each other?
The force required to double the length of a steel wire of area 1 cm2, if its Young's modulus Y= 2 × 1011/m2 is:
In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case ______.
ASSERTION (A): In an interference pattern observed in Young's double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to 1/3rd, then new fringe width remains the same.
REASON (R): Fringe width is proportional to (d/D).
How will the interference pattern in Young's double-slit experiment be affected if the screen is moved away from the plane of the slits?
How will the interference pattern in Young's double-slit experiment be affected if the source slit is moved away from the plane of the slits?
A beam of light consisting of two wavelengths 600 nm and 500 nm is used in Young's double slit experiment. The silt separation is 1.0 mm and the screen is kept 0.60 m away from the plane of the slits. Calculate:
- the distance of the second bright fringe from the central maximum for wavelength 500 nm, and
- the least distance from the central maximum where the bright fringes due to both wavelengths coincide.
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 ______.
In Young's double slit experiment the two slits are 0.6 mm distance apart. Interference pattern is observed on a screen at a distance 80 cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be ______ nm.
What is the interference pattern observed on the screen in Young's double-slit experiment?
What is the angular fringe width \(\alpha\)?
