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JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Question Bank Young's Double Slit Experiment and Biprism

  • question_answer
    In Fresnel's biprism \[(\mu =1.5)\] experiment the distance between source and biprism is 0.3 m and that between biprism and screen is 0.7m and angle of prism is \[1{}^\circ \]. The fringe width with light of wavelength\[6000\,\,{AA}\]will be [RPMT 2002]

    A) 3 cm                                        

    B) 0.011 cm

    C) 2 cm                                         

    D) 4 cm

    Correct Answer: B

    Solution :

    \[\beta =\frac{(a+b)\lambda }{2a(\mu -1)\alpha }\] where a = distance between source and biprism = 0.3 m b = distance between biprism and screen = 0.7 m. a = Angle of prism = 1°, m = 1.5, \[\lambda \]= 6000 ´ 10?10 m Hence, \[\beta =\frac{(0.3+0.7)\times 6\times {{10}^{-7}}}{2\times 0.3(1.5-1)\times ({{1}^{o}}\times \frac{\pi }{180})}\] = 1.14 ´ 10?4 m = 0.0114 cm.


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