When positioned 5 cm away from a weak point source, a photographic plate requires an exposure time of 3 s . If the plate is instead located at a distance of 10 cm from the source, the duration required to achieve an

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Quantrex Academy · Accepted Answer

Correct answer: C. 12 s

Exposure is given by the product of intensity and time,

E = I \times t

.

For a point source, intensity is inversely proportional to the square of the distance,

I \propto \dfrac{1}{r^2}

.

For identical exposure,

I_1 t_1 = I_2 t_2

.

\dfrac{t_1}{r_1^2} = \dfrac{t_2}{r_2^2}

t_2 = t_1 \left(\dfrac{r_2}{r_1}\right)^2

Substituting the given values

r_1 = 5 \text{ cm}

,

r_2 = 10 \text{ cm}

, and

t_1 = 3 \text{ s}

:

t_2 = 3 \times \left(\dfrac{10}{5}\right)^2

t_2 = 3 \times 2^2 = 12 \text{ s}

When positioned 5 cm away from a weak point source, a photographic...

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