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KVPY Sample Paper KVPY Stream-SX Model Paper-19

  • question_answer
    In Fresnel's \[\operatorname{biprism}\text{ }(\mu =1.5)\] experiment the distance between source and biprism is 0.3 m and that between biprism and screen is 0.7 m and angle of prism is \[1{}^\circ \]. The fringe width with light of wavelength \[6000\,\overset{\text{o}}{\mathop{\text{A}}}\,\] will be

    A) 3 mm

    B) 0.11 mm

    C) 2 mm   

    D) 4 mm

    Correct Answer: B

    Solution :

    \[a=0.3\operatorname{m},b=07\operatorname{m}.\,\]Angle of prism, \[A=1{}^\circ .\]
    \[\therefore \]\[D=a+b=0.3+0.7=1m.\]
    \[d=2a(\mu -1)A=2\times 0.3(1.5-1)\times \frac{\pi }{180}\]
    \[=0.0052\operatorname{m}\]
    Now \[\beta =\frac{D\lambda }{d}=\frac{1\times 6000\times {{10}^{-10}}}{0.0052}\]
    \[=1.15\times {{10}^{-4}}m=0.115\]


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