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Photometry — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsPhotometry1 Mark
Question
Light from a point source falls on a screen. If the separation between the source and the screen is increased by 1%, the illuminance will decrease (nearly) by:
  1. 0.5%
  2. 1%
  3. 2%
  4. 4%
✓

Answer

  1. 2%
Explanation:
Illuminance is given by:
$\text{E}=\frac{\text{l}_{\text{o}}\cos\theta}{\text{r}^2}$
$\theta=0^0$
$\frac{\triangle\text{r}}{\text{r}}=1\%$
$\text{E}=\frac{\text{l}_{\text{o}}}{\text{r}^2}$
Differentiating,
$\text{dE}=-2\frac{\text{l}_{\text{o}}}{\text{r}^3}\text{dr}$
As approximation differentials are replaced by $\triangle$,
$\triangle\text{E}=-2\frac{\text{l}_\text{o}}{\text{r}^2}\triangle \text{r}$
$$$\Rightarrow\triangle\text{E}=-2\frac{\text{I}_\text{o}}{\text{r}^2}\Big(\frac{\triangle\text{r}}{\text{r}}\Big)$
$\Rightarrow\triangle\text{E}=-2\text{E}\Big(\frac{\triangle\text{r}}{\text{r}}\Big)$
$\Rightarrow\frac{\triangle{\text{E}}}{\text{E}}=-2\Big(\frac{\triangle{\text{r}}}{\text{r}}\Big)$
$\Rightarrow \frac{\triangle\text{E}}{\text{E}}=-2\times1\%=-2\%$
Since, negative sign implies decrease; hence, illuminance decreases by 2%.

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