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JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Question Bank Critical Thinking

  • question_answer
    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                        

    B)            256 : 16 : 169

    C)            256 : 16 : 196                         

    D)            256 : 196 : 16

    Correct Answer: B

    Solution :

               \[{{I}_{A}}=R_{1}^{2}\] \[{{I}_{B}}={{({{R}_{1}}-{{R}_{2}})}^{2}}=R_{1}^{2}{{\left( 1-\frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{2}}=R_{1}^{2}{{\left( 1-\frac{3}{4} \right)}^{2}}=\frac{R_{1}^{2}}{16}\] \[{{I}_{C}}={{({{R}_{1}}-{{R}_{2}}+{{R}_{3}})}^{2}}\]\[=R_{1}^{2}{{\left( 1-\frac{{{R}_{2}}}{{{R}_{1}}}+\frac{{{R}_{3}}}{{{R}_{1}}} \right)}^{2}}\]                                \[=R_{1}^{2}{{\left( 1-\frac{{{R}_{2}}}{{{R}_{1}}}+\frac{{{R}_{3}}}{{{R}_{2}}}\times \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{2}}\]      \[=R_{1}^{2}{{\left( 1-\frac{3}{4}+\frac{3}{4}\times \frac{3}{4} \right)}^{2}}\]\[={{\left( \frac{13}{16} \right)}^{2}}R_{1}^{2}=\frac{169}{256}R_{1}^{2}\] \[\therefore \,\,\,{{I}_{A}}:{{I}_{B}}:{{I}_{C}}=R_{1}^{2}:\frac{R_{1}^{2}}{16}:\frac{169}{256}R_{1}^{2}=256:16:169\]


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