question_answer

In Youngs double slit experiment distance between source is 1 mm and distance between the screen and source is 1 m. If the fringe width on the screen is 0.06 cm, then \[\lambda \] is :

A)  6000\[\overset{0}{\mathop{A}}\,\]                                       

B)  4000 \[\overset{0}{\mathop{A}}\,\]

C)  1200 \[\overset{0}{\mathop{A}}\,\]                                      

D)  2400 \[\overset{0}{\mathop{A}}\,\]

Correct Answer: A

Solution :

                When distance between screen and source is D, and d the distance between coherent sources, then fringe width (W) is given by \[W=\frac{D\lambda }{d}\] where\[\lambda \]is wavelength of monochromatic light. \[\therefore \]  \[\lambda =\frac{Wd}{D}\] Given, D = 1 m,                 \[d=1\text{ }mm={{10}^{-3}}m,\]             \[W=0.06\text{ }cm,\]       \[=0.06\times {{10}^{-2}}m\] \[\therefore \]  \[\lambda =\frac{0.06\times {{10}^{-2}}\times {{10}^{-3}}}{1}\] \[=6\times {{10}^{-7}}m=6000{\AA}\]