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The colours in white light are V I B G Y O R.
Even though the colour in the middle is green, why do we take yellow to represent λavg and μavg of light? I'm really [7][7][7] and [2][2][2].
Pls tell me if You can't understand the question.
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10 Answers
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62this is easy dude... see dont get biased.. there are not 7 colors but too many of them..
lamda average is (most probably) found by taking average energy..
62it is not about equal measure..
it is like u are finding the median not the mean..
is the mean equal to the median always.. (if u remember the difference between the two from class x or ix !)
See for the data
(1,1,1,1,1,1,1,2) median is 1 but mean is 9/8
11Ok. Got it bhaiya. Thanx for helping [1]
I always keep getting silly doubts like this [2]
9thats bcos yellow is in centre of visible spectrum
if u observe carefully , red ,orang yellow are thicker bands than blue and oth
so these 3 of em dominate half of vis spec though they r 3 in no
1well, nishant bhaiya...how do we (or scientists) decide when to use mean and when median????
i mean..we cud have taken mean too....
and i'm also equally satisfied wid celestine's reply....
62average is by taking mean..
when we talk about taking average it means about mean...
what celestine is saying is also correct..
and here in this case it is a combination of 2...
if we were to find the "median" wavelength.. we would have taken highest+lowerst by 2.
1http://en.wikipedia.org/wiki/File:Linear_visible_spectrum.svg
(image link....dunno why i cudnt upload it...)
well...i don't know exactly how we calculate λavg (i'm in 11th)..
but if we do like this:
visible range = 380nm to 750nm
λavg = (380 + 750)/2 = 565nm
which is close to yellow, but inside the green boundary...
and if we calculate the mean of the average wavelengths of each colour, λavg = 548nm
which is more inside green........
(cudn't guess how we'll calculate median...)
1plz reply...am i correct???
avg does come out to be green...(although, it is quite close to yellow)