The intensity of light $I_{1}$ received by the emitter plate in Lenard's photoelectric set-up is changed by moving the source to twice the initial distance. What is the new intensity $I_{2}$ received by the emitter plate?

I₂ = I₁

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I₂ = 0.25 I₁

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I₂ = 0.5 I₁

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I₂ = 2 I₁

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Solution

The correct option is B
I₂ = 0.25 I₁

As we have seen earlier, the intensity from a point source decreases as the square of the distance from the source. If intensity $I_{1}$ was received at a distance $r_{1}, a n d I_{2} a t r_{2}$ -

$I α \frac{1}{r^{2}} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}} = \frac{{(2 r_{1})}_{1}^{2}}{r_{1}^{2}} = 4 \Rightarrow I_{2} = \frac{1}{4} I_{1} \Rightarrow I_{2} = 0.25 I_{1}$

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The intensity of light I1 received by the emitter plate in Lenard's photoelectric set-up is changed by moving the source to twice the initial distance. What is the new intensity I2 received by the emitter plate?

The correct option is B I2 = 0.25 I1 As we have seen earlier, the intensity from a point source decreases as the square of the distance from the source. If intensity I1 was received at a distance r1, and I2 at r2 - Iα1r2⇒ I1I2=r22r21=(2r1)2r21=4⇒ I2=14I1⇒ I2=0.25 I1

The intensity of light $I_{1}$ received by the emitter plate in Lenard's photoelectric set-up is changed by moving the source to twice the initial distance. What is the new intensity $I_{2}$ received by the emitter plate?