Advertisements
Advertisements
Question
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?
Options
420 nm
750 nm
630 nm
500 nm
Advertisements
Solution
420 nm
Explanation:
Position of nth bright fringe,
`y_n = (nlambdaD)/d`
Given: λ = 700 nm
∴ `y_3 = (3 xx 700 xx D)/d`
⇒ `y_3 = (2100D)/d`
The 5th bright fringe due to light of wavelength λ' is formed at y3,
∴ y5 = y3
or `(5 xx lambda^'D)/d = (2100 nmD)/d`
⇒ `lambda^' = 2100/5` = 420 nm
APPEARS IN
RELATED QUESTIONS
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.
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.
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 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 \[\left( \lambda = 700\text{ nm} \right).\]
Two balls are projected at an angle θ and (90° − θ) to the horizontal with the same speed. The ratio of their maximum vertical heights is:
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 :
In Young's double slit experiment, the distance of the 4th bright fringe from the centre of the interference pattern is 1.5 mm. The distance between the slits and the screen is 1.5 m, and the wavelength of light used is 500 nm. Calculate the distance between the two slits.
In Young's double slit experiment, show that:
`β = (λ"D")/"d"`
Where the terms have their usual meaning.
If the monochromatic source in Young’s double slit experiment is replaced by white light, then ______.
