Violet light with = 430 nm strikes a pair of slits separated by 0.505 mm . Part

Question

Violet light with = 430 nm strikes a pair of slits separated by 0.505 mm .

Part A

On a screen 1.23 m away, what's the distance between the two second-order bright fringes?

Express your answer with the appropriate units.

Part B

On a screen 1.23 m away, what's the distance between the central maximum and one of the fourth dark fringes?

Express your answer with the appropriate units.

Part C

On a screen 1.23 m away, what's the distance between the third bright fringe on one side of the central maximum and the third dark fringe on the other?

Express your answer with the appropriate units.

Explanation / Answer

given that

lambda = 430 nm = 430*10^(-9) m

d = 0.505 mm = 0.505*10^(-3) m

D = 1.23 m

part (A)

we know that

y = m*lambda*D/d

where m is the position of bright fring .

distance between the two second-order bright fringes

Y = y2 - y1

Y = ( 2*430*10^(-9)*1.23 / 0.505*10(-3) ) - ( 2*430*10^(-9)*1.23 / 0.505*10(-3) )

Y = 0

the distance between the two second-order bright fringes is zero .

part (B)

Let y is the distance from the center of the central maximum to the 4'th dark fringe,

y = (m+1/2)*lambda*D /d

y = (4+1/2)*430*10^(-9)*1.23 / 0.505*10^(-3)

y = 2618.31*10^(-6) m

y = 2.61*10^(-3) m

y = 2.61 mm

part(C)

Let the distance between the third bright fringe on one side of the central maximum and the third dark fringe on the other dide is Y

Y = Yb - Yd

Y = ( 3*430*10^(-9)*1.23 / 0.0505*10^(-3) ) - ( (3+1/2)*430*10^(-9)*1.23/ 0.505*10^(-3) )

Y = 3141.98*10^(-6) - 2094.65*10^(-6)

Y = 3.14 mm - 2.09 mm

Y = 1.05 mm

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