Calculating Light Loss and Beam Spread in Opto-Electronics Components

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SUMMARY

The discussion focuses on calculating light loss and beam spread in opto-electronics, specifically using a laser diode with a power output of 5mW for free space optical communications. It confirms that power loss occurs over distance, adhering to the inverse-square law, which states that doubling the distance results in a quarter reduction in power at the receiver. Additionally, the need for calculating beam spread is emphasized to optimize light collection by the photo-transistor, with references to Gaussian beam parameters for further understanding.

PREREQUISITES
  • Understanding of laser diode specifications and characteristics
  • Knowledge of the inverse-square law in physics
  • Familiarity with Gaussian beam theory and parameters
  • Basic principles of optical communication systems
NEXT STEPS
  • Research the application of the inverse-square law in optical systems
  • Learn about Gaussian beam propagation and its parameters
  • Explore lens design for focusing laser light onto photodetectors
  • Investigate methods for measuring light intensity over distance
USEFUL FOR

This discussion is beneficial for optical engineers, researchers in opto-electronics, and anyone involved in designing free space optical communication systems.

Gogsey
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This is not a homework question but for a project.

We are using a laser diode, one that you would normally find in a laser pointer to transfer information to a photo-transistor for a free space optical communications system.

Now I know the power output is around 5mW, but over a distance will we lose power as the light travels towards the photo-transistor? If so, how do you calculate the power at a certain distance, say 25 m or so?

Also, could someone also tell me the formula for calculating the spread of the laser light over a distance. This is so we know how much light is missing the photo-transistor and so we can correct this by using lenses to focus the light onto the photo-transistor?

Thanks in advance.
 
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Well it would still have to obey the inverse-square law. Doubling the distance from the source means a 1/4 reduction in power at the receiver end.
 

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