Advertisements
Advertisements
प्रश्न
In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.
Advertisements
उत्तर
Distance of the screen from the slits, D = 1 m
Wavelength of light used, `lambda_1` = 600 nm
Angular width of the fringe in air θ1 = 0.2°
Angular width of the fringe in water = θ2
Refractive index of water, μ = `4/3`
Refractive index is related to angular width as:
μ = `θ_1/θ_2`
θ2 = `3/4 θ_1`
= `3/4 xx 0.2`
= 0.15
Therefore, the angular width of the fringe in water will reduce to 0.15°.
APPEARS IN
संबंधित प्रश्न
Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
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?
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is a distance of 2.5 mm away from the centre. Find the width of the 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?
The separation between the consecutive dark fringes in a Young's double slit experiment is 1.0 mm. The screen is placed at a distance of 2.5m from the slits and the separation between the slits is 1.0 mm. Calculate the wavelength of light used for the experiment.
White light is used in a Young's double slit experiment. Find the minimum order of the violet fringe \[\left( \lambda = 400\text{ nm} \right)\] which overlaps with a red fringe \[\left( \lambda = 700\text{ nm} \right).\]
A plate of thickness t made of a material of refractive index µ is placed in front of one of the slits in a double slit experiment. (a) Find the change in the optical path due to introduction of the plate. (b) What should be the minimum thickness t which will make the intensity at the centre of the fringe pattern zero? Wavelength of the light used is \[\lambda.\] Neglect any absorption of light in the plate.
A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55, respectively, for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen at a distance one metre away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?
A Young's double slit apparatus has slits separated by 0⋅28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by red light \[\left( \lambda = 700\text{ nm in vacuum} \right).\] Find the fringe-width of the pattern formed on the screen.
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 interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength \[\lambda.\] Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one-fourth the maximum.
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.
What should be the path difference between two waves reaching a point for obtaining constructive interference in Young’s Double Slit experiment ?
Draw a neat labelled diagram of Young’s Double Slit experiment. Show that `beta = (lambdaD)/d` , where the terms have their usual meanings (either for bright or dark fringe).
Wavefront is ______.
Two slits in Young's interference experiment have width in the ratio 1 : 2. The ratio of intensity at the maxima and minima in their interference is ______.
Consider a two-slit interference arrangement (Figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of λ such that the first minima on the screen falls at a distance D from the centre O.
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 ______.
