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max2010maxim [7]
4 years ago
9

White light containing wavelengths from 410 nm to 750 nm falls on a grating with 7300 slits/cm. part a how wide is the first-ord

er spectrum on a screen 3.60 m away?
Physics
1 answer:
kotegsom [21]4 years ago
5 0

Answer:

Explanation:

Width of each slit d = 1 x 10⁻² / 7300

= 1.37 x 10⁻⁶ m

Distance of screen D = 3.6 m

for light of wavelength   λ   =  410 nm = 410 x 10⁻⁹ m

position of first maxima =λ D / d

= 410 x 10⁻⁹  x 3.6 / 1.37 x 10⁻⁶

= 1077.37 x 10⁻³ m

= 1.077 m

for light of wavelength   λ   =  750 nm = 750 x 10⁻⁹ m

position of first maxima =λ D / d

= 750 x 10⁻⁹  x 3.6 / 1.37 x 10⁻⁶

= 1438.54 x 10⁻³ m

= 1.438 m

width of first order spectrum = 1.438 - 1.077 m

= .361 m

36.1 cm .

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A

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Answer:

a) \Delta{t} = 5.39s

b) the motorcycle travels 155 m

Explanation:

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v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}

where:

v_{m2} is the speed of the motorcycle at time 2

v_{c} is the velocity of the car (constant)

v_{0} is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]

v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s

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x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933

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