Question

The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with different spectral distributions. Daylight lighting corresponds to the spectral distribution of the solar disk, which may be approximated as a blackbody at 5800K. Incandescent lighting from the usual household bulb corresponds approximately to the spectral distribution of a black body at 2900K. Calculate the band emission fractions for the visible region, 0.47 mu m to 0.65 mum, for each of the lighting sources. Calculate the wavelength corresponding to the maximum spectral intensity for each of the light sources

Answer 1

Answer:

a) at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

b) daylight (d) = 0.50 μm

    Incandescent ( i ) =  1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

Values are gotten from the table named: blackbody radiation functions

a) Calculate the band emission fractions for the visible region

at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

attached below is a detailed solution to the problem

b)calculate wavelength corresponding to the maximum spectral intensity

For daylight ( d ) = 2898 μm *k / 5800 k  = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm