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Q.

A small photocell is placed at a distance of $4 m$ from a photosensitive surface. When light falls on the surface the current is $5 m A$ . If the distance of cell is decreased to $1 m$ , the current will become

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a

$1.25 m A$

b

$(\frac{5}{16}) m A$

c

$80 m A$

d

$20 m A$

answer is D.

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Detailed Solution

Intensity of light $I \propto \frac{1}{d^{2}}$ , where d = distance from surface.

$\Rightarrow \frac{I_{2}}{I_{1}} = {(\frac{d_{1}}{d_{2}})}^{2} = {(\frac{4}{1})}^{2} = 16 \Rightarrow I_{2} = 16 I_{1}$

As current is proportional to intensity of light, new value of current $= 16 \times 5 mA = 80 mA$ .

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A small photocell is placed at a distance of 4m from a photosensitive surface. When light falls on the surface the current is 5mA. If the distance of cell is decreased to 1m, the current will become