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

Question

Dr. Abubacker · Accepted Answer

(b) $I_{A} = R_{1}^{2}$
$I_{B} = \left( R_{1} - R_{2} ight)^{2} = R_{1}^{2}\left( 1 - \frac{R_{2}}{R_{1}} ight)^{2} = R_{1}^{2}\left( 1 - \frac{3}{4} ight)^{2} = \frac{R_{1}^{2}}{16}$
$I_{C} = \left( R_{1} - R_{2} + R_{3} ight)^{2} = R_{1}^{2}\left( 1 - \frac{R_{2}}{R_{1}} + \frac{R_{3}}{R_{1}} ight)^{2}$
$= R_{1}^{2}\left( 1 - \frac{R_{2}}{R_{1}} + \frac{R_{3}}{R_{2}} 	imes \frac{R_{2}}{R_{1}} ight)^{2}$
$= R_{1}^{2}\left( 1 - \frac{3}{4} + \frac{3}{4} 	imes \frac{3}{4} ight)^{2} = \left( \frac{13}{16} ight)^{2}R_{1}^{2} = \frac{169}{256}R_{1}^{2}$
$	herefore I_{A}:I_{B}:I_{C} = R_{1}^{2}:\frac{R_{1}^{2}}{16}:\frac{169}{256}R_{1}^{2} = 256:16:169$

Q. 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

Explanation

Reviewed by Dr. Abubacker

Want more practice?