The intensity of light I1 received by the emitter plate in Lenard's photoel

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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?

Options:

A . I2=I1

B . I2=0.25 I1

C . I2=0.5 I1

D . I2=2 I1

Answer: Option B
:
B
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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