When the distance of a point light source from a photocell is $r_{1}$ , photoelectric current is $I_{1}$ . If the distance becomes $r_{2}$ , then the current is $I_{2}$ , The ratio ( $I_{1} : I_{2}$ ) is equal to

r_{2}^{2} : r_{1}^{2}

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r_{2} : r_{1}

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r_{1}^{2} : r_{2}^{2}

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r_{1} : r_{2}

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Solution

The correct option is A $r_{2}^{2} : r_{1}^{2}$
$I \propto \frac{1}{r^{2}}$

$\frac{I_{1}}{I_{2}} = {(\frac{r_{2}}{r_{1}})}^{2}$

$= {r_{2}}^{2} : {r_{1}}^{2}$

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Similar questions

r_{1}

r_{2}

I_{1}

I_{2}

\frac{I_{1}}{I_{2}}

If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

If the sum of the areas of two circles with radii $R_{1}$ and $R_{2}$ is equal to the area of a circle of radius R then

(a) $R_{1} + R_{2} = R$

(b) $R_{1} + R_{2} > R$

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$

R_{1}

R_{2}

R_{1} + R_{2} = R

R_{1}^{2} + R_{2}^{2} = R^{2}

R_{1} + R_{2} < R

R_{1}^{2} + R_{2}^{2} < R_{2}

\frac{{R_{1}}^{2}}{{R_{2}}^{2}} = \frac{9}{576}, \frac{R_{1}}{R_{2}} = ?

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