Physics, 23.03.2020 22:45 Mordred2143

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm.

If the intensity at the center of the central maximum is 5.0010?4W/m2 , what is the intensity at a point on the screen that is 0.830mm from the center of the central maximum?

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Answer from: dontcareanyonemo

The intensity at a point on the screen that is 0.830mm from the center of the central maximum is; I = 8.64 × 10⁻⁶ W/m²

We are given;

Wavelength; λ = 571 nm = 571 × 10⁻⁹ m

Distance of interference from screen; R = 75 cm = 0.75 m

center to center distance between slits; d = 0.640 mm = 0.00064 m

width of each slit; a = 0.434 mm = 0.000434 m

Intensity at center; I₀ = 5 × 10⁻⁴ W/m²

Distance of intensity from center of the central maximum; y = 0.83 mm = 0.00083 m

Let us first find the angle θ from the formula;

tan θ = y/R

tan θ = 0.00083/0.75

tan θ = 0.001107

θ = tan⁻¹0.001107

θ = 0.06343

θ is very small and as such we use the formula;

∅ = (2πd sinθ)/λ

∅ = (2π * 0.00064 * sin 0.06343)/(571 × 10⁻⁹)

∅ = 7.7964 rad

Similarly, for distance between the centers of slit;

β = (2πa sinθ)/λ

β = (2π * 0.000434 * sin 0.06343)/(571 × 10⁻⁹)

β = 5.287 rad

Formula for the Intensity at a point on the screen that is 0.830mm from the center of the central maximum is;

$I = I_{o}(cos^{2}\frac{\phi}{2}) [\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}}]^{2}$

$I = 5 * 10^{-4} }(cos^{2}(\frac{7.7964}{2})) [\frac{sin\frac{5.287}{2}}{\frac{5.287}{2}}]^{2}$

I = 8.64 × 10⁻⁶ W/m²

Read more about Intensity at; link

Answer from: Jasten

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

Answer from: Quest

distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

slope of a graph = change in value of y -axis / change in values of x -axis

comparing both the equations

change in value of y -axis = change in position

change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

Answer from: Quest

option number c is the right answer.

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