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Vladimir79 [104]
3 years ago
10

Monochromatic light of wavelength 687 nm is incident on a narrow slit. On a screen 1.65 m away, the distance between the second

diffraction minimum and the central maximum is 2.09 cm. (a) Calculate the angle of diffraction θ of the second minimum. (b) Find the width of the slit. Please explain with your solution
Physics
1 answer:
Sophie [7]3 years ago
3 0

Answer:

a ) 1.267 radian

b ) 1.084 10⁻³ mm

Explanation:

Distance of screen D = 1.65 m

Width of slit d = ?

Wave length of light   λ  = 687 nm.

Distance of second minimum fro centre y = 2.09 cm

Angle of diffraction = y / D

=  2.09 /1.65  

= 1.267. radian

Angle of diffraction of second minimum

= 2 λ / d

so 2 λ / d = 1.267

d = 2 λ / 1.267 = (2 x 687 ) /1.267 nm

=1084.45 nm = 1.084 x 10⁻³ mm.

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A box falls out of a stationary helicopter hovering 135 m above the ground. How long will it take to hit the ground?
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Here displacement = 135 m, Initial velocity = 0 m/s, acceleration = 9.81 m/s^2

Substituting

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3 years ago
The electric field between the plates of the cathode ray tube of an older television set can be as high as 2.5x104 N/C. Determin
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Answer:

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Parameter given:

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(a) Electric force is given (in terms of electric field) as a product of electric charge and electric field.

Mathematically:

F = qE

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(b) This electrostatic force causes the electron to accelerate with an equivalent force:

F = -ma

where m = mass of an electron

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(Note: the force is negative cos the direction of the force is opposite the direction of the electron)

Therefore:

-ma =  -4.01 * 10^{-15} N\\\\\\a = \frac{-4.01 * 10^{-15}}{-m}

Mass, m, of an electron = 9.11 * 10^{-31} kg

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The acceleration of the electron is 4.40 * 10^{15} m/s^2

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