Question

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
(a) 3 s
(b) 12 s
(c) 24 s
(d) 48 s

Answer 1

Time needed for exposure is 12 seconds.

Distance = d1 = 5cm  (Given)

Distance =  d2 = 10cm  Given)

Let the incident illuminance is = E0

Let the incident illuminance at 3cm is = E1

Let the incident illuminance at 5 cm =  E2

Thus, at 3 -  

E1 = E0 / d²1

Timee= t1 ∝ k/E1

t1 = k (5)²/E0

k/E0 = 3/25

Therefore, time required is -  

t2 = k (10)² / E0

T2 = 12

The time needed for exposure is 12 seconds.

Answer 2

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, then the time needed for the same exposure is 12 s

Explanation:

Step 1:

Given data :

$d_1$   = 5 cm = 0.05  m  

$d_2$  = 10  cm  =  0.1  m  

$t_1$ = 3 s  

$t_2$ = ?  

Let the real illumination of the incident be Eo

Let $E_d_1$ be the illuminance 3 cm away.

Let the illuminance be $E_d_2$ at 10 cm distance.

cosθ = 1

Step 2:

$t_{1} \alpha \frac{1}{E_{d 1}}$

Now,  

$t_{1} \alpha \frac{1}{E_{d 1}}$

$t_{1}=\frac{k}{E_{d 1}}$

$t_{1}=\frac{k 5^{2}}{E_{o}}$

$\frac{k}{E_{0}}=\frac{3}{25}$

Step 3:

Similarly,  

$t_{2}=\frac{k 10^{2}}{E_{0}}$

$t_{2}=\frac{3}{25} \times 10^{2}$

   $=12 s$

Therefore the correct answer is Option(b) 12 s