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Artist 52 [7]
3 years ago
5

You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. Yo

u set your slit spacing at 1.15 mm and place your screen 9.39 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.55 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength expressed in nanometers?
Physics
1 answer:
4vir4ik [10]3 years ago
7 0

Answer:

557.24nm

Explanation:

m λ = dsinθ

d = 1.01 x 10e-3   m=10

we use tanθ because of the long distance

so d (x/R) = m λ    x = distance from central fringe and R = distance from screen

(1.15x10e-3)[(4.55x10e-2)/(9.39)] = 10 λ

5.5724x10e-6 m or 557.24nm

 

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A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
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Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58

4 \Delta t^2 = 58

\Delta t^2 = 14.5

\Delta t = \sqrt{14.5} = 3.807s

(b)

s_m = 23 \Delta t + 4 \Delta t^2

s_m = 23*3.807 + 58 = 145.581 m

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