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Margaret [11]
3 years ago
15

A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maxi

mum is Io, what is the intensity 10° from the center?
Physics
1 answer:
Valentin [98]3 years ago
5 0

Answer:

The intensity at 10° from the center is 3.06 × 10⁻⁴I₀

Explanation:

The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ

I₀ = maximum intensity of light

a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m

θ = angle at intensity point = 10°

λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m

α = πasinθ/λ

= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m

= 1.0911/650 × 10³

= 0.001679 × 10³

= 1.679

Now, the intensity I is

I = I₀(sinα/α)²

= I₀(sin1.679/1.679)²

= I₀(0.0293/1.679)²

= 0.0175²I₀

= 0.0003063I₀

= 3.06 × 10⁻⁴I₀

So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀

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An LC circuit is built with a 20 mH inductor and an 8.0 PF capacitor. The capacitor voltage has its maximum value of 25 V at t =
Margaret [11]

Answer:

a) the required time is 0.6283 μs

b) the inductor current is 0.5 mA

Explanation:

Given the data in the question;

The capacitor voltage has its maximum value of 25 V at t = 0

i.e V_m = V₀ = 25 V

we determine the angular velocity;

ω = 1 / √( LC )

ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )

ω = 1 / √( 1.6 × 10⁻¹³  )

ω = 1 / 0.0000004

ω = 2.5 × 10⁶ s⁻¹

a) How much time does it take until the capacitor is fully discharged for the first time?

V_m =  V₀sin( ωt )

we substitute

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

divide both sides by 25 V

sin( 2.5 × 10⁶ × t ) = 1

( 2.5 × 10⁶ × t ) = π/2

t = 1.570796 / (2.5 × 10⁶)

t = 0.6283 × 10⁻⁶ s

t = 0.6283 μs

Therefore, the required time is 0.6283 μs

b) What is the inductor current at that time?

I(t) = V₀√(C/L) sin(ωt)

{ sin(ωt) = 1 )

I(t) = V₀√(C/L)

we substitute

I(t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )

I(t) = 25 × 0.00002

I(t) = 0.0005 A

I(t) = 0.5 mA

Therefore, the inductor current is 0.5 mA

8 0
3 years ago
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