Question

In a grease spot photometer, light from a lamp dirty with a chimney is exactly balanced by a point source distant $10\,{\text{cm}}$ from the grease spot. On clearing the chimney, the point source is moved $2\,{\text{cm}}$ to obtain balance again. The percentage of light absorbed by the dirty chimney is nearly
A. $56\%$
B. $44\%$
C. $36\%$
D. $64\%$

Answer 1

Use the formula for intensity of the light at a point away from the source of light. This formula gives the relation between the intensity of the light, power output of the source of light and distance from the source of light. First calculate the change distance of the lamp. Derive the relation between the intensity of light and the distance from the source. Use the formula for decrease in intensity of the light in terms of distance from the source and calculate the required answer.

Formula used:
The intensity $I$ of light at point is given by
$I = \dfrac{P}{{4\pi {R^2}}}$ …… (1)
Here, $P$ is the power output of the source of light and $R$ is the distance of the source from the point where intensity is to be measured.

Complete step by step answer:
We have given that the initial distance of the light source from the chimney is $10\,{\text{cm}}$ to obtain the balance.
${R_1} = 10\,{\text{cm}}$
We have also given that the distance of the light source is changed by $2\,{\text{cm}}$ to obtain the balance again.
Hence, the new distance ${R_2}$ of the light source from chimney is
${R_2} = {R_1} - \left( {2\,{\text{cm}}} \right)$
Substitute $10\,{\text{cm}}$ for ${R_1}$ in the above equation.
${R_2} = \left( {10\,{\text{cm}}} \right) - \left( {2\,{\text{cm}}} \right)$
$\Rightarrow {R_2} = 8\,{\text{cm}}$
Therefore, the new distance of the lamp from the chimney is $8\,{\text{cm}}$.

From equation (1), we can conclude that the intensity $I$ of the light at a point away from the source is directly proportional to the square of the distance $R$ of the point where the intensity is to be measured from the source of light which is lamp as the power $P$ of the source of light which is lamp is constant.
$I \propto \dfrac{1}{{{R^2}}}$
Hence, the equation for decrease in the intensity of the light from the lamp is given by
$\% \,{\text{decrease}} = \dfrac{{R_1^2 - R_2^2}}{{R_1^2}} \times 100$

Substitute $10\,{\text{cm}}$ for ${R_1}$ and $8\,{\text{cm}}$ for ${R_2}$ in the above equation.
$\% \,{\text{decrease}} = \dfrac{{{{\left( {10\,{\text{cm}}} \right)}^2} - {{\left( {8\,{\text{cm}}} \right)}^2}}}{{{{\left( {10\,{\text{cm}}} \right)}^2}}} \times 100$
$\Rightarrow \% \,{\text{decrease}} = \dfrac{{100 - 64}}{{100}} \times 100$
$\therefore \% \,{\text{decrease}} = 36\%$
Thus, the decrease in the intensity of the light is $36\%$. Therefore, the intensity of the light absorbed by the dirty chimney is $36\%$.

Hence, the correct option is C.

Note: The students may think why we have decreased and not increased the changed distance by 2 cm. The students should keep in mind that some of the intensity of the light from the lamp is absorbed by the chimney and the intensity of the light is more for the smaller distance as intensity is inversely proportional to the square of distance.