In a grease spot photometer,light from a lamp | vedclass

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(C) Let the luminous intensity of the lamp be $I_1$ and that of the point source be $I_2$. Let the distance of the lamp from the grease spot be $x$.In

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(C) Let the luminous intensity of the lamp be $I_1$ and that of the point source be $I_2$. Let the distance of the lamp from the grease spot be $x$.In the first case (dirty chimney),the intensity of the lamp is $I_1' = I_1(1 - a)$,where $a$ is the fraction of light absorbed. The balance condition is given by:$\frac{I_1'}{x^2} = \frac{I_2}{10^2} \implies \frac{I_1'}{I_2} = \frac{x^2}{100} \quad (1)$In the second case (clean chimney),the intensity of the lamp is $I_1$. The point source is moved by $2 \, cm$ to obtain balance. Since the lamp is now brighter,the point source must be moved further away to $10 + 2 = 12 \, cm$ (or closer if the balance shifts,but based on the diagram,the distance becomes $8 \, cm$ as the lamp is now brighter). The balance condition is:$\frac{I_1}{x^2} = \frac{I_2}{8^2} \implies \frac{I_1}{I_2} = \frac{x^2}{64} \quad (2)$Dividing equation $(1)$ by $(2)$:$\frac{I_1'}{I_1} = \frac{x^2/100}{x^2/64} = \frac{64}{100} = 0.64$This means $I_1' = 0.64 \, I_1$. The fraction of light transmitted is $0.64$,so the fraction absorbed is $1 - 0.64 = 0.36$.Therefore,the percentage of light absorbed is $36\%$.

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