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Alexandra [31]
3 years ago
9

Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit.

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm. What is the width of the slit?
Physics
1 answer:
Effectus [21]3 years ago
7 0

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, \lambda=608\ nm=608\times 10^{-9}\ m

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

y=\dfrac{mD\lambda}{a}

where

a = width of the slit

a=\dfrac{mD\lambda}{y}

a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}

a = 0.000167 m

a=1.67\times 10^{-4}\ m

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

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3 years ago
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<h3>What are values?</h3>
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<h3>What are some instances of values?</h3>
  • Spirituality.
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1 year ago
A passenger bus in Zurich, Switzerland derived its motive power from the energy stored in a large flywheel. The wheel was brough
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Answer:

18695271.3209 J

989.167794757 s

Explanation:

r = Radius = \dfrac{1.7}{2}=0.85\ m

N = 2970 rpm

m = Mass of flywheel = 1070 kg

P = Power = 1.89\times 10^4\ W

Moment of inertia

I=\dfrac{1}{2}mr^2\\\Rightarrow I=\dfrac{1}{2}1070\times 0.85^2\\\Rightarrow I=386.5375\ kgm^2

Angular speed

\omega=2970\times \dfrac{2\pi}{60}\\\Rightarrow \omega=311.017672705\ rad/s

Kinetic energy

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}386.5375\times 311.017672705^2\\\Rightarrow K=18695271.3209\ J

The kinetic energy is 18695271.3209 J

Time is given by

t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{18695271.3209}{1.89\times 10^4}\\\Rightarrow t=989.167794757\ s

The time taken is 989.167794757 s

4 0
3 years ago
An electron is accelerated through 2400 V from rest and then enters a region in which there is a uniform 1.70 T magnetic field.
aalyn [17]

Answer:

Explanation:

Let v be the velocity acquired by electron in electric field

V q = 1/2 m v²

V is potential difference applied on charge q , m is mass of charge , v is velocity acquired

2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²

v² = 844 x 10¹²

v = 29.05 x 10⁶ m /s

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= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶

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