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(B) Let the amplitudes of the first,second,and third $HPZ$ be $R_1, R_2, R_3$ respectively. Given the ratio of consecutive amplitudes is $R_2/R_1 = R_

Vedclass · Accepted Answer

$256 : 16 : 169$

Vedclass · Answer

(B) Let the amplitudes of the first,second,and third $HPZ$ be $R_1, R_2, R_3$ respectively. Given the ratio of consecutive amplitudes is $R_2/R_1 = R_3/R_2 = 3/4$.At point $A$ (one $HPZ$): $I_A = R_1^2$.At point $B$ (two $HPZ$): $I_B = (R_1 - R_2)^2 = R_1^2(1 - R_2/R_1)^2 = R_1^2(1 - 3/4)^2 = R_1^2(1/4)^2 = R_1^2/16$.At point $C$ (three $HPZ$): $I_C = (R_1 - R_2 + R_3)^2 = R_1^2(1 - R_2/R_1 + R_3/R_1)^2 = R_1^2(1 - 3/4 + (3/4)^2)^2 = R_1^2(1 - 3/4 + 9/16)^2 = R_1^2((16 - 12 + 9)/16)^2 = R_1^2(13/16)^2 = (169/256)R_1^2$.Thus,$I_A : I_B : I_C = R_1^2 : (R_1^2/16) : (169/256)R_1^2 = 256 : 16 : 169$.

$A$ wavefront presents one,two and three $HPZ$ at points $A, B$ and $C$ respectively. If the ratio of consecutive amplitudes of $HPZ$ is $4 : 3$,then the ratio of resultant intensities at these points will be

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