question_answer

         In young's double - slit experiment, the separation between the slits is d, distance between the slit and screen is D (D>>d). In the interference pattern, there is a maxima exactly in front of each slit. Then the possible wavelength(s) used in the experiment are        

A) \[{{d}^{2}}/D,\,{{d}^{2}}/2D,\,{{d}^{2}}/3D\]

B) \[{{d}^{2}}/D,\,{{d}^{2}}/3D,\,{{d}^{2}}/5D\]

C) \[{{d}^{2}}/2D,\,{{d}^{2}}/4D,\,{{d}^{2}}/6D\]

D) None of these

Correct Answer: C

Solution :

[c] \[{{S}_{2}}P-{{S}_{1}}P=\frac{dy}{D}=\frac{d\times (d/2)}{D}=\frac{{{d}^{2}}}{2D}\] \[\frac{{{d}^{2}}}{2D}=n\lambda \] \[\lambda =\frac{d}{2nD}\], n=1, 2,       ??.. \[\lambda =\frac{{{d}^{2}}}{2D},\,\frac{{{d}^{2}}}{4D},\frac{{{d}^{2}}}{6D}\]