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The separatio...

Question

The separation between a point source of light anda screen is increased by 1.5%. The percentage change in the intensity is

Increases by 3%
Decreases by 3%
Increases by 0.75%
Decreases by 0.75%

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Solution

Dear Student,

$I n t e n s i t y i s i n v e r s e l y p r o p o r t i o n a l t o s q u a r e o f t h e d i s \tan c e ., I = \frac{K}{r^{2}} I' = \frac{K}{{(1.015 r)}^{2}} = \frac{K}{1.030225 r^{2}} a s w e s e e t h e i n t e n s i t y d e c r e a s e s . p e r c e n t a g e d e c r e a s e i n i n t e n s i t y = \frac{I - I'}{I} = \frac{\frac{K}{r^{2}} - \frac{K}{1.030225 r^{2}}}{\frac{K}{r^{2}}} * 100 = 2.93 p e r c e n t t h a t i s a p p r o x 3 p e r c e n t e d e c r e a s e . R e g a r d s$

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