To determine the duration of the \(E_2\) dose at 50 cm, we need to understand the inverse square law, which applies to the intensity of light. The law states that intensity is inversely proportional to the square of the distance from the source. Thus, if the distance is halved, the intensity increases by a factor of four.
\(I_1 = \frac{Dose}{(100)^2}\)
\(I_2 = \frac{Dose}{(50)^2}\)
\(I_1 \times t_1 = I_2 \times t_2\)
\(\frac{1}{(100)^2} \times 60 = \frac{1}{(50)^2} \times t_2\)
\(\frac{60}{10000} = \frac{t_2}{2500}\)
\(t_2 = \frac{60 \times 2500}{10000}\)
\(t_2 = 15 \times 2.5 = 37.5\) seconds
Thus, the correct answer is 37.5 seconds.