A light, flashing regularly, consists of cycles, each cycle having a dark phase and a light phase. The frequency of this light is measured in cycles per second. As the frequency is increased, the eye initially perceives a series of flashes of light, then a coarse flicker, a fine flicker, and ultimately a steady light. The frequency at which the flickering disappears is called the fusion frequency. The table below shows the results of an experiment in which the fusion frequency F was measured as a function of the light intensity

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(a) Here , clearly it has been given that we have to find the model of the type F = a In(I) + b , using the first and last terms of the data . F = a In(I) + b     ...........(1) So , First term of the data is - When I = 0.8 , F = 8  And last term of the data is - When I = 980 , F = 38.2  Now using these two cases in equation (1) , we get - 8=a \ln (0.8)+b     ,and  38.2=a \ln (980)+b Here , In (0.8) = - 0.2231435513 and , In (980) = 6.8875525717 Thus , we get - 8=a(-0.2231435513)+b \cdots(2) 38.2=a(6.8875525717)+b \cdots(3) Now , sunbtracting equation (2) from (3)  , we get - 30.2=a(7.110696123) \Rightarrow a=\frac{30.2}{7.110696123}=4.2471228523   Now , using a = 4.2471228523 in equation (2) , we get - 8=(4.2471228523)(-0.2231435513)+b \Rightarrow 8=-0.9477180761+b \Rightarrow b=8+0.9477180761=8.9477180761 i.e.,  a=4.25 and  b=8.95 Now , using these vales in equation (1) , we get the required logarithmic model for the data as - F=4.25 \ln (I)+8.95 (b) We have F=4.25 \ln (I)+8.95 Now , we have to find the frequency(F) of fusion when intensity (I) is equal to 48.4 Thus using I = 48.4 in above equation , we get - F=4.25 \ln (48.4)+8.95 \Rightarrow F=4.25(3.8794998137)+8.95 \Rightarrow F=16.4878742083+8.95 \Rightarrow F=25.4378742083 i.e.,  F=25.4 Hence , we get that , For an intensity of 48.4 ,  The model predicts a frequency of 25.4 as compared to the observed frequency of 25.3   (c) We have - F=4.25 \ln (I)+8.95 We have to find the value of Intensity (I) , when frequency(F) is equal to 28.3 Thus , using F = 28.3 in above equation , we get  28.3=4.25 \ln (I)+8.95 \Rightarrow 4.25 \ln (I)+8.95=28.3 \Rightarrow 4.25 \ln (I)=28.3-8.95 \Rightarrow 4.25 \ln (I)=19.35 \Rightarrow \ln (I)=\frac{19.35}{4.25} \Rightarrow \ln (I)=4.5529411765 \Rightarrow I=e^{4.5529411765} \Rightarrow I=94.9111486417 i.e.,  I=94.9 Hence , we get that , For Frequency (F) = 28.3 ,  The model predicts an intensity of 94.9 as compared to the observed intensity of 92.5  ...