JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Morning)

  • question_answer
    In a Young's double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle \[\frac{1}{40}\] rad by using light of wavelength\[{{\lambda }_{1}}\]. When the light of wavelength \[{{\lambda }_{2}}\] is used a bright fringe is seen at the same angle in the same set up. Given that \[{{\lambda }_{1}}\] and \[{{\lambda }_{2}}\] are in visible range (380 nm to 740 nm), their values are - [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

    A) 400 nm, 500 nm                        

    B) 625 nm, 500 nm

    C) 380 nm, 500 nm            

    D)   380 nm, 525 nm

    Correct Answer: B

    Solution :

    \[d\text{ }sin\text{ }\theta =n\lambda ,\] In question \[\theta =\frac{1}{40}\] which is very small \[\therefore \text{ }sin\text{ }\theta =\theta =\frac{1}{40}\] \[d\times \theta =n\lambda ,\] \[n\,\,=\,\,\frac{d\theta }{\lambda }\,\,=\,\,\frac{0.1\,mm}{40\,\lambda }\] \[n=\frac{2500}{\lambda }nm\,\] When \[\lambda =380\] \[{{n}_{1}}\,=\,\frac{2500}{350}\,\,=\,\,6.578\] \[for\,\,\lambda =740\,nm\] \[{{n}_{2}}\,=\,\frac{2500}{740}\,=\,3.378\] \[\therefore \,\,\,\,n=4,\,\,5,\,\,6\] \[for\text{ }n=4,\text{ }\lambda =625\text{ }nm\] \[for\text{ }n=5,\text{ }\lambda =500\text{ }nm\]


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