NEET-XII-Physics
22: Photometry
- #9A point source emitting 628 lumen of luminous flux uniformly in all directions is placed at the origin. Calculate the illuminance on a small area placed at (1.0 m, 0, 0) in such a way that the normal to the area makes an angle of 37° with the X-axis.Ans : Given,
Luminous flux = 628 lumen
Angle made by the normal with the x axis (`` \theta ``) = 37ₒ
Distance of point, r = 1 m
Since the radiant flux is distributed uniformly in all directions, the solid angle will be 4`` \,\mathrm{\,\pi \,}``.
∴ Luminous intensity, l =`` \frac{\,\mathrm{\,Luminous\,}\,\mathrm{\,Flux\,}}{\,\mathrm{\,Solid\,}\,\mathrm{\,angle\,}}``
`` ``
`` =\frac{628}{4\,\mathrm{\,\pi \,}}=50\,\mathrm{\,candela\,}``
`` ``
`` I\,\mathrm{\,lluminance\,}\left(E\right)\,\mathrm{\,is\,}\,\mathrm{\,given\,}\,\mathrm{\,by\,},``
`` E=l\,\mathrm{\,cos\,}\frac{\theta }{{r}^{2}}``
On substituting the respective values we get,
`` E=50\times \frac{\,\mathrm{\,cos\,}37°}{{1}^{2}}``
`` =50\times \frac{(4/5)}{1}=40\,\mathrm{\,lux\,}``
So, the illuminance on the area is 40 lux.
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