In young double slit experiment if the distance between two slits is halved and distance between the slits and the screen is doubled then what will be the effect on fringe width ?
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20
Hey Mate
Here's your answer
In the YDSE experiment of Wave Optics the fringe width of the interference is D(lambda)/d
where D is distance between slit and screen and d is distance between slits
On doubling D it becomes 2D(lambda)/d and on taking half of d it becomes 2D(lambda)/(d/2) which on doing inverse becomes 4D(lambda)/d
Hence on doubling D and taking half of d ..... the fringe width becomes 4times the original
Hope this helps
Be brainly
Here's your answer
In the YDSE experiment of Wave Optics the fringe width of the interference is D(lambda)/d
where D is distance between slit and screen and d is distance between slits
On doubling D it becomes 2D(lambda)/d and on taking half of d it becomes 2D(lambda)/(d/2) which on doing inverse becomes 4D(lambda)/d
Hence on doubling D and taking half of d ..... the fringe width becomes 4times the original
Hope this helps
Be brainly
Answered by
8
Answer:
Explanation:
Separation between the slit = 1/2
Distance between the slits and screen = 2x
Young’s double slit experiment on interference of light has the formula -
X = λD/d
where X is the band width or fringe width, λ is the wave length of the monochromatic light, D is the distance between the slits and screen, and d is the distance between the slits.
Maxima = ( n+1) λD/d
Yn = Distance of nth
Maxima = nλD/d
Therefore, D2 = 2D1
= D2/D1 × d1/d2
= 2×2
= 4
Thus, If the distance between the slits is doubled, the fringr width is is quadrupled.
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