A movie projector forms an image 3.5 m long | vedclass

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(B) The illuminance $I$ is defined as the luminous flux per unit area,$I = \frac{\Phi}{A}$.Since the total luminous flux $\Phi$ remains constant (assu

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$10^4:1$

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(B) The illuminance $I$ is defined as the luminous flux per unit area,$I = \frac{\Phi}{A}$.Since the total luminous flux $\Phi$ remains constant (assuming negligible absorption),the illuminance is inversely proportional to the area of the image,$I \propto \frac{1}{A}$.Since the area $A$ is proportional to the square of the linear dimensions $(A \propto L^2)$,we have $I \propto \frac{1}{L^2}$.Therefore,the ratio of illuminance on the slide $(I_s)$ to the illuminance on the screen $(I_i)$ is given by:$\frac{I_s}{I_i} = \frac{L_i^2}{L_s^2} = \left( \frac{L_i}{L_s} \right)^2$Given $L_i = 3.5 \ m = 3500 \ mm$ and $L_s = 35 \ mm$:$\frac{I_s}{I_i} = \left( \frac{3500 \ mm}{35 \ mm} \right)^2 = (100)^2 = 10000 = 10^4:1$.

$A$ movie projector forms an image $3.5 \ m$ long of an object $35 \ mm$ long. Supposing there is negligible absorption of light by the aperture,what is the ratio of the illuminance on the slide to the illuminance on the screen?

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