Photometry — Physics STD 11 Science — Question

3. 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%.