Question

Light from a point source falls on a screen. If the separation between the source and the screen is increased by 1%, the illuminance will decrease (nearly) by
(a) 0.5%
(b) 1%
(c) 2%
(d) 4%

Answer 1

I think The answer is option (b) 1℅

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Answer 2

Light from a point source falls on a screen. If the separation between the source and the screen is increased by 1%, the illuminance will decrease (nearly) by 2%

Explanation:

Step 1:

As we know that

Illuminance is given through:

$E=\frac{I_{0} \cos \theta}{r^{2}}$

$\theta=0$

$\frac{\Delta r}{r}=1 \%$

$E=\frac{I_{0}}{r^{2}}$

Step 2:

On Differentiating the above equation, we get the following

 $d E=-2 \frac{I_{0}}{r^{3}} d r$

Step 3:

As approximation differentials are replaced by Δ,

$\Delta E=-2 \frac{I_{0}}{r^{3}} \Delta r$

$\Delta E=-2 \frac{I_{0}}{r^{2}}\left(\frac{\Delta r}{r}\right)$

$\Delta E=-2 E\left(\frac{\Delta r}{r}\right)$

$\frac{\Delta E}{E}=-2\left(\frac{\Delta r}{r}\right)$

$\frac{\Delta E}{E}=-2 \times 1 \%$

$\frac{\Delta E}{E}=-2 \%$

Since negative sign implies a reduction; therefore, illuminance decreases by 2 %.