A wavefront presents one, two and three HPZ at points A, B and C … - Vidyadip

MCQ

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 point will be

A
$169 : 16 : 256$
✓
$256 : 16 : 169$
C
$256 : 16 : 196$
D
$256 : 196 : 16$

✓

Answer

Correct option: B.

$256 : 16 : 169$

b
(b) ${I_A} = R_1^2$
${I_B} = {({R_1} - {R_2})^2} = R_1^2{\left( {1 - \frac{{{R_2}}}{{{R_1}}}} \right)^2} = R_1^2{\left( {1 - \frac{3}{4}} \right)^2} = \frac{{R_1^2}}{{16}}$
${I_C} = {({R_1} - {R_2} + {R_3})^2}$$ = R_1^2{\left( {1 - \frac{{{R_2}}}{{{R_1}}} + \frac{{{R_3}}}{{{R_1}}}} \right)^2}$
$ = R_1^2{\left( {1 - \frac{{{R_2}}}{{{R_1}}} + \frac{{{R_3}}}{{{R_2}}} \times \frac{{{R_2}}}{{{R_1}}}} \right)^2}$
$ = R_1^2{\left( {1 - \frac{3}{4} + \frac{3}{4} \times \frac{3}{4}} \right)^2}$$ = {\left( {\frac{{13}}{{16}}} \right)^2}R_1^2 = \frac{{169}}{{256}}R_1^2$
$\therefore \,\,\,{I_A}:{I_B}:{I_C} = R_1^2:\frac{{R_1^2}}{{16}}:\frac{{169}}{{256}}R_1^2 = 256:16:169$

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