Quantum Properties of Light - Revision history

Gespitia3: /* Computational Model */

2022-04-27T04:28:24Z

Computational Model

← Older revision		Revision as of 00:28, 27 April 2022
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	==Computational Model==		==Computational Model==
	For a model concerned with the quantum nature of light in terms of the double slit experiment, please access this website: ~~<a href = "~~https://ophysics.com/l4.html~~"></a>~~		For a model concerned with the quantum nature of light in terms of the double slit experiment, please access this website: https://ophysics.com/l4.html

	==Examples==		==Examples==

2022-04-27T04:28:03Z

Computational Model

← Older revision		Revision as of 00:28, 27 April 2022
Line 57:		Line 57:

	==Computational Model==		==Computational Model==
	For a model concerned with the quantum nature of light in terms of the double slit experiment, please access this website: https://ophysics.com/l4.html		For a model concerned with the quantum nature of light in terms of the double slit experiment, please access this website: <a href = "https://ophysics.com/l4.html"></a>

	==Examples==		==Examples==

2022-04-27T04:27:13Z

Computational Model

← Older revision		Revision as of 00:27, 27 April 2022
Line 57:		Line 57:

	==Computational Model==		==Computational Model==
			For a model concerned with the quantum nature of light in terms of the double slit experiment, please access this website: https://ophysics.com/l4.html

	==Examples==		==Examples==

2022-04-24T22:16:15Z

Solution

← Older revision		Revision as of 18:16, 24 April 2022
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	<math>\frac{1}{\lambda_{e}^{2}} = \frac{1}{\lambda^{2}} + \frac{1}{\lambda^{'2}} - \frac{2cos\theta}{\lambda\lambda^{'}}</math>		<math>\frac{1}{\lambda_{e}^{2}} = \frac{1}{\lambda^{2}} + \frac{1}{\lambda^{'2}} - \frac{2cos\theta}{\lambda\lambda^{'}}</math>

			Lastly, we must make the appropiate subsitution using the first simpliflied equation above. This proof is left as an exercise to the reader.

	==Connectedness==		==Connectedness==

2022-04-24T22:13:58Z

Solution

← Older revision		Revision as of 18:13, 24 April 2022
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	where we have written <math>p_{e} = \frac{h}{\lambda_{e}}</math>		where we have written <math>p_{e} = \frac{h}{\lambda_{e}}</math>

			Next, we eliminate <math>\phi</math> and <math>\lambda_{e}</math> from these equations. Eliminating <math>\phi</math> from the last two equations yields

			<math>\frac{1}{\lambda_{e}^{2}} = \frac{1}{\lambda^{2}} + \frac{1}{\lambda^{'2}} - \frac{2cos\theta}{\lambda\lambda^{'}}</math>

	==Connectedness==		==Connectedness==

2022-04-24T22:11:05Z

Solution

← Older revision		Revision as of 18:11, 24 April 2022
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	<math>\frac{1}{\lambda} = \frac{cos\theta}{\lambda^{'}} + \frac{cos\phi}{\lambda_{e}}</math>		<math>\frac{1}{\lambda} = \frac{cos\theta}{\lambda^{'}} + \frac{cos\phi}{\lambda_{e}}</math>

	\frac{sin\theta}{\lambda^{'}} = \frac{sin\phi}{\lambda_{e}}		<math>\frac{sin\theta}{\lambda^{'}} = \frac{sin\phi}{\lambda_{e}}</math>

	where we have written <math>p_{e} = \frac{h}{\lambda_{e}}</math>		where we have written <math>p_{e} = \frac{h}{\lambda_{e}}</math>

2022-04-24T22:10:20Z

Solution

← Older revision		Revision as of 18:10, 24 April 2022
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	<math>\frac{1}{\lambda} = \frac{cos\theta}{\lambda^{'}} + \frac{cos\phi}{\lambda_{e}}</math>		<math>\frac{1}{\lambda} = \frac{cos\theta}{\lambda^{'}} + \frac{cos\phi}{\lambda_{e}}</math>

			\frac{sin\theta}{\lambda^{'}} = \frac{sin\phi}{\lambda_{e}}

			where we have written <math>p_{e} = \frac{h}{\lambda_{e}}</math>

	==Connectedness==		==Connectedness==

2022-04-24T22:07:57Z

Solution

← Older revision		Revision as of 18:07, 24 April 2022
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	<math>\frac{1}{\lambda} + \frac{1}{\lambda_{c}} = \frac{1}{\lambda^{'}} + \sqrt{\frac{1}{\lambda^{2}_{e}} + \frac{1}{\lambda^{2}_{c}}}</math>		<math>\frac{1}{\lambda} + \frac{1}{\lambda_{c}} = \frac{1}{\lambda^{'}} + \sqrt{\frac{1}{\lambda^{2}_{e}} + \frac{1}{\lambda^{2}_{c}}}</math>

			<math>\frac{1}{\lambda} = \frac{cos\theta}{\lambda^{'}} + \frac{cos\phi}{\lambda_{e}}</math>

	==Connectedness==		==Connectedness==

2022-04-24T22:05:19Z

Solution

← Older revision		Revision as of 18:05, 24 April 2022
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	Simpliflying the above equations,		Simpliflying the above equations,

	<math>\frac{1}{\lambda} + \frac{1}{\lambda_{c}} = \frac{1}{\lambda^{'}} + \sqrt{\frac{1}{\lambda^{2}_{e}} + \frac{1}{\lambda^{2}_{c}}</math>		<math>\frac{1}{\lambda} + \frac{1}{\lambda_{c}} = \frac{1}{\lambda^{'}} + \sqrt{\frac{1}{\lambda^{2}_{e}} + \frac{1}{\lambda^{2}_{c}}}</math>

	==Connectedness==		==Connectedness==

2022-04-24T22:05:08Z

Solution

← Older revision		Revision as of 18:05, 24 April 2022
Line 118:		Line 118:
	Simpliflying the above equations,		Simpliflying the above equations,

	<math>\frac{1}{\lambda} + \frac{1}{\lambda_{c}} = \frac{1}{\lambda^{'}} + \sqrt{\frac{1}{\lambda^{2}_{e}} + \frac{1}{\lambda^{2}_}{c}}</math>		<math>\frac{1}{\lambda} + \frac{1}{\lambda_{c}} = \frac{1}{\lambda^{'}} + \sqrt{\frac{1}{\lambda^{2}_{e}} + \frac{1}{\lambda^{2}_{c}}</math>

	==Connectedness==		==Connectedness==