NEET-XII-Physics
22: Photometry
- #5Light 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%digAnsr: CAns : Correct option
(c)
​Illuminance is given by:
`` E=\frac{{I}_{o}\,\mathrm{\,cos\,}\theta }{{r}^{2}}``
`` ``
`` \theta ={0}^{0}``
`` \frac{\Delta r}{r}=1\%``
`` E=\frac{{I}_{o}}{{r}^{2}}``
`` \,\mathrm{\,Differentiating\,},``
`` dE=-2\frac{{I}_{o}}{{r}^{3}}dr``
`` \,\mathrm{\,As\,}\,\mathrm{\,approximation\,}\,\mathrm{\,differentials\,}\,\mathrm{\,are\,}\,\mathrm{\,replaced\,}\,\mathrm{\,by\,}\,\mathrm{\,\Delta \,},``
`` \,\mathrm{\,\Delta E\,}=-2\frac{{\,\mathrm{\,I\,}}_{\,\mathrm{\,o\,}}}{{\,\mathrm{\,r\,}}^{3}}\,\mathrm{\,\Delta r\,}``
`` \Rightarrow \,\mathrm{\,\Delta E\,}=-2\frac{{\,\mathrm{\,I\,}}_{\,\mathrm{\,o\,}}}{{\,\mathrm{\,r\,}}^{2}}\left(\frac{\,\mathrm{\,\Delta r\,}}{\,\mathrm{\,r\,}}\right)``
`` \Rightarrow \,\mathrm{\,\Delta E\,}=-2\,\mathrm{\,E\,}\left(\frac{\,\mathrm{\,\Delta r\,}}{\,\mathrm{\,r\,}}\right)``
`` \Rightarrow \frac{\,\mathrm{\,\Delta E\,}}{\,\mathrm{\,E\,}}=-2\left(\frac{\,\mathrm{\,\Delta r\,}}{\,\mathrm{\,r\,}}\right)``
`` \Rightarrow \frac{\,\mathrm{\,\Delta E\,}}{\,\mathrm{\,E\,}}=-2\times 1\%=-2\%``
Since negative sign implies decrease hence illuminance decrease by 2 %.
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