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Anton [14]
4 years ago
10

Using a 685 nm wavelength laser, you form the diffraction pattern of a 1.11 mm wide slit on a screen. You measure on the screen

that the 11th dark fringe is 9.85 cm away from the center of the central maximum. How far is the screen located from the slit
Physics
1 answer:
velikii [3]4 years ago
5 0

Answer:

13.8 m

Explanation:

Dark fringes are formed in a single slit experiment due to destructive interference that occurs due to interference.

The position of these dark fringes formed on a screen is given by:

y = \frac{\lambda }{d} (m + 1/2)D

where y = position of  mth minimum

m = order of the minimum

D = distance of the slit from the screen

d = width of the slit

λ = wavelength of the light used

We need to find D:

D = \frac{yd} {\lambda (m + 1/2)} \\

From the question:

m = 11

y = 9.85 cm = 0.0985 m

λ  = 6.83 * 10^{-7} m

d = 1.11 mm = 0.0011 m

Therefore:

D = \frac{0.0985 *0.0011} {6.83 * 10^{-7} *(11 + 1/2)} \\\\D = \frac{0.00010835} {6.83 * 10^{-7}  * (23/2)} \\\\D = 13.8 m

The slit is 13.8 m far from the screen

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\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s

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\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

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