Problem from Intro to Health Physics by Cember and Johnson

[xyz_physics]

Hello guys, can you please help me figure out how to solve this problem. Many thanks!

A scanning He–Ne laser that scans at a rate of 10 s−1 emits 5 mW through an
aperture of 0.7 cm. If the beam divergence is 5 milliradian, then, for an intrabeamviewing
distance of 200 cm, calculate
(a) the time during each scan that the pupil of the eye can be exposed,
(b) the radiant exposure per scan,
(c) the average irradiance at the cornea, and
(d) the hazard class that should be assigned to this laser?

 

[Monique]

You should post your question in the https://www.physicsforums.com/forumdisplay.php?f=152" and follow the guidelines that are listed when you post the new thread.

 

[quicknote]

Sure, I would be happy to help you solve this problem. First, let's define some variables:
- N = scanning rate of laser (10 s^-1)
- P = power of laser (5 mW)
- D = beam divergence (5 milliradian)
- d = aperture of laser (0.7 cm)
- L = intrabeam viewing distance (200 cm)

(a) To calculate the time during each scan that the pupil of the eye can be exposed, we can use the formula: time = distance/speed. In this case, the distance is the diameter of the beam at the viewing distance, which can be calculated using the formula: diameter = 2 x distance x tan(angle). So, the diameter of the beam at 200 cm is 2 x 200 cm x tan(5 milliradian) = 0.35 cm. The speed is the scanning rate, which is 10 s^-1. Therefore, the time during each scan that the pupil of the eye can be exposed is 0.35 cm / 10 s^-1 = 0.035 s.

(b) To calculate the radiant exposure per scan, we can use the formula: radiant exposure = power x time. The time we calculated in part (a) is in seconds, so we need to convert it to milliseconds (ms) to match the units of power (mW). 0.035 s = 35 ms. Therefore, the radiant exposure per scan is 5 mW x 35 ms = 175 mJ.

(c) To calculate the average irradiance at the cornea, we can use the formula: irradiance = power / area. The area of the beam at the cornea can be calculated using the formula: area = π x (diameter/2)^2. So, the area of the beam at the cornea is π x (0.35 cm / 2)^2 = 0.096 cm^2. Therefore, the average irradiance at the cornea is 5 mW / 0.096 cm^2 = 52.1 mW/cm^2.

(d) To determine the hazard class that should be assigned to this laser, we can use the American National Standards Institute (ANSI) classification system. Based on the calculations in parts (a) and (b), the radiant exposure per scan is less than 1 J/cm^2, which falls