question_answer

In Young's 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 K is

A)  \[6000\text{ }\overset{\text{o}}{\mathop{\text{A}}}\,\]      

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

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

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

Correct Answer: A

Solution :

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