NEET Sample Paper NEET Sample Test Paper-90

  • question_answer
    The Fraunhoffer ?diffraction? pattern of a single slit is formed in the focal plane of a lens of focal length 1 m. The width of slit is 0.3 mm. If third minimum is formed at a distance of 5 mm from central maximum, then wavelength of light will be

    A) \[5000\,\overset{{}^\circ }{\mathop{\,A}}\,\]     

    B)        \[2500\,\,\overset{{}^\circ }{\mathop{A}}\,\]

    C) \[7500\,\,\overset{{}^\circ }{\mathop{A}}\,\]     

    D)         \[8500\,\,\overset{{}^\circ }{\mathop{A}}\,\]

    Correct Answer: A

    Solution :

    [a] a \[\sin \theta =n\lambda \] \[\frac{a\,\,x}{f}=3\,\lambda \] (since \[\theta \] is very small so \[\sin \theta \approx \tan \theta \approx \theta =x/f\]) or \[\lambda =\frac{a\,x}{3f}=\frac{0.3\times {{10}^{-3}}\times 5\times {{10}^{-3}}}{3\times 1}\] \[=5\times {{10}^{-7}}m=5000\overset{{}^\circ }{\mathop{A}}\,.\]


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