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jenyasd209 [6]
3 years ago
8

Light from two lasers is incident on an opaque barrier with a single slit of width 4.0 x 10^-4 m. One laser emits light of wavel

ength 480 nm and the other is 640 nm. A screen to view the light intensity pattern is 2.0 m behind the barrier. The distance from the center of the pattern to the nearest completely dark spot (dark for both colors) is ____ cm. (include 2 digits after the decimal point)
Physics
1 answer:
Sergio [31]3 years ago
7 0

Answer:

a) y = 2.4 x 10⁻³ m = 0.24 cm

b) y = 3.2 x 10⁻³ m = 0.32 cm

Explanation:

The formula of Young's Double Slit experiment will be used here:

y = \frac{\lambda L}{d}\\\\

where,

y = distance between dark spots = ?

λ = wavelength

L = distance of screen = 2 m

d = slit width = 4 x 10⁻⁴ m

a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:

y = \frac{(4.8\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}

<u>y = 2.4 x 10⁻³ m = 0.24 cm</u>

<u></u>

a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:

y = \frac{(6.4\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}

<u>y = 3.2 x 10⁻³ m = 0.32 cm</u>

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Answer:

a) t=2s

b) h_{max}=1024ft

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Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

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To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

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h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

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Answer:

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Answer:

(a) False

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