In an experiment to investigate the relationship between light intensity I lux on a surface and distance d metres from a light source, a student obtained the following results. One of the values of I is incorrect. Table:
d (metres)25102030
I (lux)875141354.223.89
It is known that II and dd are related by the equation I=AdnI = Ad^{n}, where AA and nn are constants.

  1. (a)

    Plot lg⁡I\lg I against lg⁡d\lg d and draw a straight line graph.

  2. (b)

    Use your graph to

    1. (i)

      identify the incorrect reading of II and estimate its correct value.

    2. (ii)

      estimate the value of AA and of nn.

  3. (c)

    Using the table of values, explain, with working, how the light intensity II changes when the distance dd from the light source is doubled.

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Worked solution

7(a)

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

Taking logarithms to base 10 for each value:

dd25102030
lg⁡d\lg d0.300.300.700.701.001.001.301.301.481.48
II875141354.223.89
lg⁡I\lg I2.942.942.152.151.541.540.630.630.590.59

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

Plot the points (lg⁡d,lg⁡I)(\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\lg d = 1.30.

Answer

A straight line graph plotted with lg⁡d\lg d on the xx-axis and lg⁡I\lg I on the yy-axis, passing through the points (0.30,2.94)(0.30, 2.94), (0.70,2.15)(0.70, 2.15), (1.00,1.54)(1.00, 1.54), and (1.48,0.59)(1.48, 0.59), with the point (1.30,0.63)(1.30, 0.63) visibly off the line.

7(b)(i)

1. Identify the incorrect reading from the plotted points

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

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

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

3. Calculate the correct value of $I$

lg⁡I=0.942\lg I = 0.942
I=100.942≈8.75I = 10^{0.942} \approx 8.75

Answer

Incorrect reading: 4.224.22
Correct value: 8.758.75

7(b)(ii)

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

I=AdnI = Ad^n
lg⁡I=lg⁡(Adn)\lg I = \lg(Ad^n)
lg⁡I=nlg⁡d+lg⁡A\lg I = n\lg d + \lg A

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

Comparing lg⁡I=nlg⁡d+lg⁡A\lg I = n\lg d + \lg A with Y=mX+cY = mX + c:
Gradient m=nm = n
YY-intercept c=lg⁡Ac = \lg A

3. Calculate the gradient to estimate $n$

Using points from the line of best fit, e.g., (0.30,2.94)(0.30, 2.94) and (1.00,1.54)(1.00, 1.54):
n=1.54−2.941.00−0.30=−1.400.70=−2n = \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)(1.00, 1.54) and n=−2n = -2:
1.54=−2(1.00)+lg⁡A1.54 = -2(1.00) + \lg A
lg⁡A=3.54\lg A = 3.54
A=103.54≈3467≈3500A = 10^{3.54} \approx 3467 \approx 3500 (to 2 s.f.)

Answer

A≈3500A \approx 3500, n≈−2n \approx -2

7(c)

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

From the table, when d=5d = 5, I=141I = 141.
When d=10d = 10 (which is double of 55), I=35I = 35.

2. Calculate the ratio of the corresponding light intensities

Ratio of I=35141≈0.248≈14I = \frac{35}{141} \approx 0.248 \approx \frac{1}{4}

Answer

The light intensity becomes approximately one quarter of its original value.

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