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AMU Medical AMU Solved Paper-2006

  • question_answer
    The first diffraction minimum due to single slit diffraction is 9, for a light of wavelength \[5000\overset{\text{o}}{\mathop{\text{A}}}\,\]. It the width of the slit is \[\text{1}\times \text{1}{{0}^{-\text{4}}}\text{ cm},\] then the value of 9 is

    A) \[30{}^\circ \]                     

    B) \[45{}^\circ \]

    C) \[60{}^\circ \]

    D) \[15{}^\circ \]

    Correct Answer: A

    Solution :

                     The distance of first diffraction minimum from the central principal maximum                 \[\beta =\frac{D\lambda }{d}\]                 \[\frac{\beta }{D}=\frac{\lambda }{d}\] or            \[d=\frac{\lambda }{\sin \theta }\] \[\Rightarrow \]               \[\sin \theta =\frac{\lambda }{d}=\frac{5000\times {{10}^{-8}}}{1\times {{10}^{-4}}}\]                 \[\sin \theta =0.5=sin{{30}^{o}}\]                 \[\theta ={{30}^{o}}\]


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