As light penetrates a slab of glass, the rate of loss of intensity of the light with respect to the depth of the glass is proportional to the intensity at that depth. If the light loses 5 per cent of its intensity in penetrating 10 per cent of the glass, what percentage is lost in penetrating the slab?
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Is this a logistics problem, I'm unsure on how to start
- Thread starter esloveday
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D
Deleted member 4993
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esloveday said:As light penetrates a slab of glass, the rate of loss of intensity of the light with respect to the depth of the glass is proportional to the intensity at that depth. If the light loses 5 per cent of its intensity in penetrating 10 per cent of the glass, what percentage is lost in penetrating the slab?
\(\displaystyle \frac{dI}{dh} \, = \, -k * I\)
Now continue...
Please show your work, indicating exactly where you are stuck - so that we know where to begin to help you
D
Deleted member 4993
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esloveday said:lnI=-kh+c
I=e[sup:mstdrp4y]-kh+c[/sup:mstdrp4y]
I = I[sub:mstdrp4y]o[/sub:mstdrp4y] *e[sup:mstdrp4y]-kh[/sup:mstdrp4y]
at h/ho = 0.10 .......................we get I/I[sub:mstdrp4y]o[/sub:mstdrp4y] = 0.95
0.95 = *e[sup:mstdrp4y]-k* h[sub:mstdrp4y]o[/sub:mstdrp4y] * 0.1[/sup:mstdrp4y]
k = ln (0.95)/(-h[sub:mstdrp4y]o[/sub:mstdrp4y]] * 0.1)
then at h = h[sub:mstdrp4y]o[/sub:mstdrp4y]
I/I[sub:mstdrp4y]o[/sub:mstdrp4y] = e[sup:mstdrp4y]-ln (0.95)/(-h[sub:mstdrp4y]o[/sub:mstdrp4y] * 0.1)* h[sub:mstdrp4y]o[/sub:mstdrp4y][/sup:mstdrp4y] = e[sup:mstdrp4y]ln (0.95)/( 0.1)[/sup:mstdrp4y] ...... then cotinue....
0.95I=e[sup:mstdrp4y]-k(0.1h)+c[/sup:mstdrp4y] which I'm guessing is wrong as it yields no useful solution .... only if you refused to see with eyes open
I/I[sub:3uajq9a4]0[/sub:3uajq9a4]=0.60 approxI/I[sub:3uajq9a4]0[/sub:3uajq9a4] = e[sup:3uajq9a4]-ln (0.95)/(-h[sub:3uajq9a4]o[/sub:3uajq9a4] * 0.1)* h[sub:3uajq9a4]o[/sub:3uajq9a4][/sup:3uajq9a4]= e[sup:3uajq9a4]ln (0.95)/( 0.1)[/sup:3uajq9a4]
so loses about 40% of initial intensity when penetrating slab