Advertisements
Advertisements
Question
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).
Options
Both A and R are true and R is the correct explanation of A
Both A and R are true and R is NOT the correct explanation of A
A is true but R is false
A is false and R is also false
Advertisements
Solution
A is true but R is false
APPEARS IN
RELATED QUESTIONS
In Young' s experiment the ratio of intensity at the maxima and minima . in the interference pattern is 36 : 16. What is the ratio of the widths of the two slits?
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.
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?
A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?
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.
When a beam of light is used to determine the position of an object, the maximum accuracy is achieved, if the light is ______.
In Young's double slit experiment, the minimum amplitude is obtained when the phase difference of super-imposing waves is: (where n = 1, 2, 3, ...)
Two slits, 4mm apart, are illuminated by light of wavelength 6000 A° what will be the fringe width on a screen placed 2 m from the slits?
Why is the diffraction of sound waves more evident in daily experience than that of light wave?
Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (λ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :
