is it half stops or 2 stops?

[foxxkat]

hi, confused with what i was told from different shops.

50mm f1.8 vs 50mm f1.4

some say it's only half a stop difference (difference of 0.4)

some say it's TWO stops difference.

who is right?

 

[ExplorerZ]

hi, confused with what i was told from different shops.

50mm f1.8 vs 50mm f1.4

some say it's only half a stop difference (difference of 0.4)

some say it's TWO stops difference.

who is right?

2.5stop of f1.8 would be f0.8.. yes f1.8 > f1.4 is only about 2/3 stop

 

[grantyale]

2*log2(F1/F2)

It's never F1-F2

 

[Clockunder]

1 stop more = double the amount of light
1 stop less = half the amount of light

Area of circle = pie * r^2
Area of circle twice the size above = 2 * pie * r^2
= pie * [(square root of 2) * r]^2

So the radius and diameter of the bigger circle is (square root of 2) of the radius and diameter of the smaller circle which is half the size of the bigger circle.

In other words, the radius and diameter of the smaller circle is 1/(square root of 2) of the radius and diameter of the bigger cicle which is twice the size of the smaller circle..

square root of 2 = 1.4142

In photography, F number is the diaphragm divided by the focal length.

For the same focal length, doubling the size of the diaphragm will let in double the amount of light = 1 stop more.

Although the diaphragm is not exactly a circle, the concept of area is the same.

Doubling the diaphragm will increase the radius/diameter by a factor of (square root of 2) or 1.4142.

Halving the diaphgragm will decrease the radius and diamenter by a factor of 1/(square root of 2 or 1.4142)

So the stops for F numbers are :
F/1
F/(1 * 1.4142) = F/1.4 which is 1 stop less than F/1
F/(1 * 1.4142 * 1.4142) = F/(1.4142)^2 = F/2 which is 2 stops less than F/1 and 1 stop less than F/1.4
F/(1* 1.4142 * 1.4142 * 1.4142) = F/(1.4142^3) = F/2.8
And so on……..

When F/1 and F/2 is only 2 stops difference, F/1.4 and F/1.8 can’t possibly be 2.5 stops difference.

In general :
1 stop difference = 1.4142
2 stop difference = 1.4142^2
3 stop difference = 1.4142^3
Each power difference of (square root of 2) is 1 stop.

To calculate the no. of stops between 2 F numbers :

Let A and B be the 2 F numbers and n be the no. of stops.

A = B * (1.4142)^n

(1.4142)^n = A/B

Log (1.4142)^n = log (A/B) ………………can use any logarithm base, 10 or natural or etc.

n * log 1.4142 = log (A/B)

n = [log(A/B)/(log 1.4142)]

General equation : n = (log F1/F2) / log (square root of 2)

Since square root of 2 = 2^0.5,

n = log (F1/F2) / log 2^0.5

= log (F1/F2) / 0.5 log 2

= 2 * log (F1/F2) / log 2

= 2 * lg (F1/F2) where lg is log base 2 (This is the equation given by grantyale).

The easier way on calculator is just use any log base with the equation
n = log (F1/F2) / log (square root of 2)

You may use 1.4 as square root of 2 for approximation because the F number we see are also rounded up/down figures.

For F/1.4 and F/1.8,

1.8 = (1.4)^n

n = log(1.8/1.4)/log 1.4

= 0.75

So F/1.8 is 0.75 stop less than F/1.4.

You may also use the same logic to calculate what the F number is given another F number and their no. of stop difference.

For e.g. what is the F number which is 2.5 stops more light than F/1.8?
A= B * (1.4)^n

1.8 = (B) * (1.4)^2.5

B = 1.8/(1.4)^2.5 = 0.78 = approximately 0.8

So F/0.8 is 2.5 stops brighter than F/1.8.

or

A = 1.8 * (1.4)^2.5

= 4.2

So F/4.2 is 2.5 stops less than F/1.8.

(No surprise since there is already 2 stops difference between F/2 and F/4).

 

[airfins]

:sweat::sweat::sweat: very chim leh... all the formula and the solving, but must say its a very detailed and step by step explanation :thumbsup: Reminds me of my school daysss...Thank God its over liao :bsmilie:

 

[chalib]

1 stop more = double the amount of light
1 stop less = half the amount of light

Area of circle = pie * r^2
Area of circle twice the size above = 2 * pie * r^2
= pie * [(square root of 2) * r]^2

So the radius and diameter of the bigger circle is (square root of 2) of the radius and diameter of the smaller circle which is half the size of the bigger circle.

In other words, the radius and diameter of the smaller circle is 1/(square root of 2) of the radius and diameter of the bigger cicle which is twice the size of the smaller circle..

square root of 2 = 1.4142

In photography, F number is the diaphragm divided by the focal length.

For the same focal length, doubling the size of the diaphragm will let in double the amount of light = 1 stop more.

Although the diaphragm is not exactly a circle, the concept of area is the same.

Doubling the diaphragm will increase the radius/diameter by a factor of (square root of 2) or 1.4142.

Halving the diaphgragm will decrease the radius and diamenter by a factor of 1/(square root of 2 or 1.4142)

So the stops for F numbers are :
F/1
F/(1 * 1.4142) = F/1.4 which is 1 stop less than F/1
F/(1 * 1.4142 * 1.4142) = F/(1.4142)^2 = F/2 which is 2 stops less than F/1 and 1 stop less than F/1.4
F/(1* 1.4142 * 1.4142 * 1.4142) = F/(1.4142^3) = F/2.8
And so on……..

When F/1 and F/2 is only 2 stops difference, F/1.4 and F/1.8 can’t possibly be 2.5 stops difference.

In general :
1 stop difference = 1.4142
2 stop difference = 1.4142^2
3 stop difference = 1.4142^3
Each power difference of (square root of 2) is 1 stop.

To calculate the no. of stops between 2 F numbers :

Let A and B be the 2 F numbers and n be the no. of stops.

A = B * (1.4142)^n

(1.4142)^n = A/B

Log (1.4142)^n = log (A/B) ………………can use any logarithm base, 10 or natural or etc.

n * log 1.4142 = log (A/B)

n = [log(A/B)/(log 1.4142)]

General equation : n = (log F1/F2) / log (square root of 2)

Since square root of 2 = 2^0.5,

n = log (F1/F2) / log 2^0.5

= log (F1/F2) / 0.5 log 2

= 2 * log (F1/F2) / log 2

= 2 * lg (F1/F2) where lg is log base 2 (This is the equation given by grantyale).

The easier way on calculator is just use any log base with the equation
n = log (F1/F2) / log (square root of 2)

You may use 1.4 as square root of 2 for approximation because the F number we see are also rounded up/down figures.

For F/1.4 and F/1.8,

1.8 = (1.4)^n

n = log(1.8/1.4)/log 1.4

= 0.75

So F/1.8 is 0.75 stop less than F/1.4.

You may also use the same logic to calculate what the F number is given another F number and their no. of stop difference.

For e.g. what is the F number which is 2.5 stops more light than F/1.8?
A= B * (1.4)^n

1.8 = (B) * (1.4)^2.5

B = 1.8/(1.4)^2.5 = 0.78 = approximately 0.8

So F/0.8 is 2.5 stops brighter than F/1.8.

or

A = 1.8 * (1.4)^2.5

= 4.2

So F/4.2 is 2.5 stops less than F/1.8.

(No surprise since there is already 2 stops difference between F/2 and F/4).

wonderful explanation dude.... :bigeyes: you are not major in math? ;p

 

[foxxkat]

holysh1t.. i thot it's a simple question but Einstein came out to explain :sweat::sweatsm:

very detailed and 'easy' to understand. :thumbsup:
thanks!

 

[zoossh]

f/1.4, f/2.0, f/2.8, f/4, f/5.6, f/8, f/11, f/16, f/22

< 1 stop difference, abt 2/3.
2 stop difference is f/2.8

 

[sapphy]

Very nice. Llearnt something while reading this thread =]