A light filter is a plate of thickness `d` whose absorption coefficient depends on wavelength `lambda` as
`x(lambda) = alpha (1-lambda//lambda_(0))^(2) cm^(-1)`,
where `alpha` and `lambda_(0)` are constants. Find the passband `Delta lambda` of this light filter, that is the band at whose edge the attenuation of light is `eta` times that at the wavelength `lambda_(0)`. The coefficient of reflection from teh surfaces of the light filter is assumed to be the same at all wavelengths.

Explanation

At the wavelength `lambda_(0)` the absotption coeffcient vanishes and loss intransmission is entirely due to reflection. This factor is the same at all wavelengths and therefore cancels out in calculating the pass bank and we need not worrt about it. Now
`T_(0) =` (transmissivity al `lambda = lambda_(0)) = (1-rho)^(2)`
`T =` transmissivity at `lambda = (1 -rho)^(2) e^(-chi(lambda)d)`
The edges of the passband are `lambda_(0) +- (Delta lambda)/(2)` and at the edge
`(T)/(T_(0)) = e^(-alphad) ((Deltalambda)/(2lambda_(0)))^(2) = eta`
Thus `(delta lambda)/(2 lambda_(0)) = sqrt((In(1)/(eta))//propd)`
or `Delta lambda = 2lambda_(0) sqrt((1)/(alphad)(In(1)/(eta)))`