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const2013 [10]
3 years ago
7

Blue light (λ = 475 nm) is sent through a single slit with a width of 2.1 µm. What is the maximum possible number of bright frin

ges, including the central maximum, produced on the screen? (Hint: What is the largest angle that can be used?)
Physics
1 answer:
Alexxandr [17]3 years ago
5 0

Answer:

  m = 4

Explanation:

The expression that explains the constructive interference of a diffraction pattern is

         a sin θ = m λ

where a  is the width of the slit and λ the wavelength

         sin θ = m λ / a

The maximum value is for when the sine is 1, let's substitute

         1 = m λ/a  

         m = a /λ

let's reduce the magnitudes to the SI system

        a = 2.1 um = 2.1 10⁻⁶

        lam = 475 nm = 475 10⁻⁹ m

let's calculate

        m = 2.1  10⁻⁶ / 475 10⁻⁹

        m = 4.42

with m must be an integer the highest value is

         m = 4

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Answer:

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Explanation:

By conservation of momentum, we have

\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.

Writing for each direction at a time,

m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.

Their kinetic energy changed by

K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) =  \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2  + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.

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