在探究表面上的光照强度 I（勒克斯）与距离光源的距离 d（米）之间关系的实验中，一名学生得到了以下结果。其中一个 I 的值是错误的。 表格：

d (米)	2	5	10	20	30
I (勒克斯)	875	141	35	4.22	3.89

已知 $I$ 和 $d$ 满足关系式 $I = Ad^{n}$，其中 $A$ 和 $n$ 为常数。

1. Calculate the values of $\lg d$ and $\lg I$ for each data point.

Taking logarithms to base 10 for each value:

2. Plot the points and draw the line of best fit.

Plot the points $(\lg d, \lg I)$ on a Cartesian plane. Draw a straight line of best fit that passes through the majority of the points, ignoring the anomalous point at $\lg d = 1.30$.

1. Identify the incorrect reading from the plotted points

From the graph of $\lg I$ against $\lg d$, the point corresponding to $d = 20$ (where $\lg d \approx 1.30$) lies significantly off the line of best fit.
Therefore, the incorrect reading is $I = 4.22$.

2. Estimate the correct value of $\lg I$ from the line of best fit

At $\lg d = \lg 20 \approx 1.301$, the corresponding value on the line of best fit is $\lg I \approx 0.942$.

3. Calculate the correct value of $I$

$\lg I = 0.942$
$I = 10^{0.942} \approx 8.75$

1. Convert the non-linear equation to linear form using logarithms

$I = Ad^n$
$\lg I = \lg(Ad^n)$
$\lg I = n\lg d + \lg A$

2. Identify the gradient and y-intercept from the linear form

Comparing $\lg I = n\lg d + \lg A$ with $Y = mX + c$:
Gradient $m = n$
$Y$-intercept $c = \lg A$

3. Calculate the gradient to estimate $n$

Using points from the line of best fit, e.g., $(0.30, 2.94)$ and $(1.00, 1.54)$:
$n = \frac{1.54 - 2.94}{1.00 - 0.30} = \frac{-1.40}{0.70} = -2$

4. Calculate the y-intercept to estimate $A$

Using the point $(1.00, 1.54)$ and $n = -2$:
$1.54 = -2(1.00) + \lg A$
$\lg A = 3.54$
$A = 10^{3.54} \approx 3467 \approx 3500$ (to 2 s.f.)

1. Identify a pair of data points from the table where the distance is doubled

From the table, when $d = 5$, $I = 141$.
When $d = 10$ (which is double of $5$), $I = 35$.

2. Calculate the ratio of the corresponding light intensities

Ratio of $I = \frac{35}{141} \approx 0.248 \approx \frac{1}{4}$