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Likurg_2 [28]
2 years ago
14

Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wavelengt

h. It then falls on two slits separated by 0.490 mm . In the resulting interference pattern on a screen 2.12 m away, adjacent bright fringes are separated by 2.86 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Determining wavelength. Part A What is the wavelength of the light that falls on the slits
Physics
1 answer:
s2008m [1.1K]2 years ago
4 0

Answer:

λ = 6.61 x 10⁻⁷ m = 661 nm

Explanation:

From the Young's Double Slit experiment, the the spacing between adjacent bright or dark fringes is given by the following formula:

Δx = λL/d

where,

Δx = fringe spacing = 2.86 mm = 2.86 x ⁻³ m

L = Distance between slits and screen = 2.12 m

d = slit separation = 0.49 mm = 0.49 x 10⁻³ m

λ = wavelength of light = ?

Therefore,

2.86 x 10⁻³ m = λ(2.12 m)/(0.49 x 10⁻³ m)

(2.86 x 10⁻³ m)(0.49 x 10⁻³ m)/(2.12 m) = λ

<u>λ = 6.61 x 10⁻⁷ m = 661 nm</u>

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Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be
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The period will change a 0,036 % relative to its initial state

Explanation:

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since there was no torque ( no rotational force applied)

ζ=0 → Δ(Iω)=0 → I₂ω₂ -I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ =ω₁/ω₂ , (2) represents final state and (1) initial state

we know also that ω=2π/T , where T is the period of the pendulum

I₂/I₁ =ω₁/ω₂ = (2π/T₁)/(2π/T₂)= T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁² , I₂ = 1/3 ML₂²  → I₂/I₁ = (1/3 ML₂²)/(1/3 ML₁²)= L₂²/L₁²= (L₂/L₁)²

since

L₂= L₁ (1+αΔT) , L₂/L₁=1+αΔT  , where ΔT is the change in temperature

now putting all together

T₂/T₁=I₂/I₁=(L₂/L₁)² = (1+αΔT) ²

finally

%change in period =(T₂-T₁)/T₁ = T₂/T₁ - 1 = (1+αΔT) ² -1

%change in period =(1+αΔT) ² -1 =[ 1+18×10⁻⁶ °C⁻¹ *10 °C]² -1 = 3,6 ×10⁻⁴ = 3,6 ×10⁻² %  = 0,036 %

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3 years ago
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