MHB What fraction of light do these glasses transmit

  • Thread starter Thread starter fxacx
  • Start date Start date
  • Tags Tags
    Fraction Light
Click For Summary
Welders utilize a logarithmic scale to determine the shade number of protective eyewear based on the fraction of light transmitted, expressed in the equation n = 1 − (7logT)/3. For a welder using glasses that transmit ⅛ of the light, the calculation yields a shade number of approximately 6.5. Viewing a solar eclipse through #14 welding glasses is deemed safe, and the corresponding fraction of light transmitted can be calculated using the rearranged equation, resulting in T being approximately 0.0002. The discussion emphasizes the importance of understanding the logarithmic relationship in selecting appropriate protective eyewear. Accurate calculations are crucial for ensuring safety in welding and during solar events.
fxacx
Messages
4
Reaction score
0
Welders use a logarithmic scale to identify protective eyewear. The shade number n, is given by the
equation n = 1 − 7logT/3 , where T is the fraction of visible light that glass transmits.

a. What shade number should a welder use that only transmits ⅛ of the light entering the glass?
b Viewing a solar eclipse through #14 welding glasses is considered safe. What fraction of light do
these glasses transmit.
 
Mathematics news on Phys.org
fxacx said:
Welders use a logarithmic scale to identify protective eyewear. The shade number n, is given by the
equation n = 1 − 7logT/3 , where T is the fraction of visible light that glass transmits.

a. What shade number should a welder use that only transmits ⅛ of the light entering the glass?
b Viewing a solar eclipse through #14 welding glasses is considered safe. What fraction of light do
these glasses transmit.

is the equation $n=1-7\log \left(\dfrac{T}{3}\right)$ ... ?

please confirm
 

Attachments

  • yarab.JPG
    yarab.JPG
    1.5 KB · Views: 126
fxacx said:
Welders use a logarithmic scale to identify protective eyewear. The shade number n, is given by the
equation n = 1 − (7log(T))/3 , where T is the fraction of visible light that glass transmits.

a. What shade number should a welder use that only transmits ⅛ of the light entering the glass?
So T= 1/8. n= 1- (7 log(1/8))/3. Can you do the arithmetic?

b. Viewing a solar eclipse through #14 welding glasses is considered safe. What fraction of light do these glasses transmit.
Now n= 14 so 14= 1- (7 log(T))/3. 14- 1= 13= -(7/3) log(T).
-(3/7)(13)= -39/7= log(T). Do you know how to "reverse" log?
 
Last edited by a moderator:
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 10 ·
Replies
10
Views
13K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
28K
Replies
5
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
2
Views
3K