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avanturin [10]
3 years ago
9

Using a 679 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t

hat the 14 th dark fringe is 8.01 cm away from the center of the central maximum. How far is the screen located from the slit?
Physics
1 answer:
sveta [45]3 years ago
7 0

To solve the problem, apply the concepts related to constructive and destructive interference. For this purpose, we will start by defining the distance from the dark fringe from the central maximum in terms of the slit separation, the number of fringe the wavelength and the distance of the screen to the slit. Mathematically this is,

y_n = \frac{(2n-1)\lambda D}{2d}

D = \frac{2dy_n}{(2n-1)\lambda}

Here,

d = Slit separation

n = Number of fringe

y_n = Distance of the dark fringe from the central maximum

D = Distance of the screen to the slit

\lambda = Wavelength

Our values are given as,

\lambda = 679nm = 679*10^{-9}m

d = 1.1mm = 1.1*10^{-3}m

y_{14} = 8.01cm = 8.01*10^{-2} m

n = 14

Replacing,

D = \frac{2(1.1*10^{-3})(8.01*10^{-2})}{(2(14)-1)(679*10^{-9})}

D = 9.61m

Therefore the screen is located from the slit to 9.61m

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A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately
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Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

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acceleration of police car = 10 mi/hr = 4.47 m/s²

V_{f}  = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

V_{f} =  V_{i} + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁  = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁  will be

d_{c} = V_{c} × t₁

we substitute

d_{c} = 22.352 × 7

d_{c}  = 156.46 m

now distance between polive car and speeding car

Δd =  d_{c} - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( V_{f} - V_{c} )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = V_{f} × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

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