An electric bulb of luminous intensity I is suspended at a height h from the center of the table having a circular surface diameter $2 r$ , the illuminace at the center of the circular disc will be

\frac{I}{r^{2}}

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\frac{I}{r}

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\frac{I}{h^{2}}

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\frac{I}{h}

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Solution

The correct option is C $\frac{I}{h^{2}}$
Illuminance E at the surface, distance x away is $E = I / x^{2}$
here as the bulb is h distance away from the bulb. $E = I / h^{2}$

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An electric bulb of luminous intensity I is suspended at a height h from the center of the table having a circular surface diameter 2r, the illuminace at the center of the circular disc will be

The correct option is C Ih2Illuminance E at the surface, distance x away is E=I/x2here as the bulb is h distance away from the bulb. E=I/h2

An electric bulb of luminous intensity I is suspended at a height h from the center of the table having a circular surface diameter $2 r$ , the illuminace at the center of the circular disc will be

The correct option is C $\frac{I}{h^{2}}$
Illuminance E at the surface, distance x away is $E = I / x^{2}$
here as the bulb is h distance away from the bulb. $E = I / h^{2}$