A photographic plate placed a distance of 5 cm from a weak point source is exposed for 3 s. If the plate is kept at a distance of 10 cm from the source, the time needed for the same exposure is _____________ .

#### Options

3s

12s

24s

48s

#### Solution

12s

Here,

\[d_1 = 5 cm = 0 . 05 m\]

\[ d_2 = 10 cm = 0 . 1 m\]

\[ t_1 = 3 s\]

\[ t_2 = ?\]

Let the actual incident illuminance be \[E_o\]

Let the iluminance at 3cm distance be \[E_{d1}\]

Let the iluminance at 10cm distance be \[E_{d2} \]

\[\cos\theta = 1\]

\[ E_{d1} = \frac{E_o}{{d_1}^2}\]

Now,

\[ t_1 \alpha\frac{1}{E_{d1}}\]

\[ \Rightarrow t_1 = \frac{k}{E_{d1}}\]

\[ \Rightarrow t_1 = \frac{k 5^2}{E_o}\]

\[ \Rightarrow \frac{k}{E_o} = \frac{3}{25}\]

Similarly,

\[ \Rightarrow t_2 = \frac{k {10}^2}{E_o}\]

\[ \Rightarrow t_2 = \frac{3}{25} \times {10}^2 = 12 s\]