2-1: please help explain why C is correct You are a distance "d" away from asource emitting light waves ofamplitude “A". If you move away todistance “4d", what amplitude of waveswill you receive there?A. amplitude = AB. amplitude = A/16C. amplitude = A/4%3DD. amplitude = A/2E amplitude =2A

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2-1: please help explain why C is correct

You are a distance "d" away from a
source emitting light waves of
amplitude “A". If you move away to
distance “4d", what amplitude of waves
will you receive there?
A. amplitude = A
B. amplitude = A/16
C. amplitude = A/4
%3D
D. amplitude = A/2
E amplitude =2A

Expert Solution

Step 1

if A is the amplitude of the wave, then the intensity I can be given as

$A^{2} α I A m p l i t u d e s q u a r e i s p r o p o r t i o n a l t o t h e i n t e n s i t y I α \frac{1}{d^{2}} (d i s t h e d i s \tan c e f r o m t h e s o u r c e t o t h e o b s e r v a t i o n p o i n t) i n t e n s i t y o f t h e w a v e a t a n y p o i n t i s i n v e r s e l y p r o p o r t i o n a l t o t h e s q u a r e o f t h e d i s \tan c e o f s e p e r a t i o n I = \frac{k}{d^{2}}, k i s t h e p r o p o r t i o n a l i t y c o n s \tan t \frac{I_{1}}{I_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}} d_{1} = d, d_{2} = 4 d (g i v e n i n t h i s p r o b l e m) \frac{I_{1}}{I_{2}} = \frac{16 d^{2}}{d^{2}} = 16$

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You are a distance "d" away from a source emitting light waves of amplitude “A". If you move away to distance "4d", what amplitude of waves will you receive there?