What fraction of light do these glasses transmit

In summary: So T= antilog(-39/7). (Or raise 10 to the (-39/7) power)In summary, welders use a logarithmic scale to identify protective eyewear, with the shade number n being determined by the equation n = 1 − (7log(T))/3, where T is the fraction of visible light that glass transmits. For a welder who wants to only transmit ⅛ of the light entering the glass, they should use a shade number of 1- (7 log(1/8))/3. Viewing a solar eclipse through #14 welding glasses is considered safe, which means that these glasses transmit a fraction of light determined by the equation T= antilog(-39
  • #1
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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.
 
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  • #2
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
 
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  • #4
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?
 
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