A) 0.65 cm
B) 4.8 mm
C) 1.63 mm
D) 3.25 cm
Correct Answer: C
Solution :
The fringe width for nth bright fringe is given by \[{{\beta }_{n}}=n\frac{D\lambda }{d}\] where d is distance between coherent sources, D the distance between screen and source. For n = 5 \[{{\beta }_{5}}=5\frac{D\lambda }{2d}\] For nth dark fringe, the fringe width is given by \[{{\beta }_{n}}=(2n-1)\frac{D\lambda }{2D}\] For third dark fringe \[{{\beta }_{3}}=(6-1)\frac{D\lambda }{2d}=\frac{5D\lambda }{2\,d}\] \[\therefore \] \[{{\beta }_{5}}-{{\beta }_{3}}=\frac{5D\lambda }{d}-\frac{5}{2}\frac{D\lambda }{d}\] \[{{\beta }_{5}}-{{\beta }_{3}}=\frac{5}{2}\frac{D\lambda }{d}\] Given, \[\lambda =6.5\times {{10}^{-7}}m\], D = 1 m, \[d=1\,mm=1\times {{10}^{-3}}m\] \[\therefore \] \[{{\beta }_{5}}-{{\beta }_{3}}=\frac{5\times 1\times 6.5\times {{10}^{-7}}}{2\times {{10}^{-3}}}\] \[{{\beta }_{5}}-{{\beta }_{4}}=1.625\times {{10}^{-3}}m\] \[{{\beta }_{5}}-{{\beta }_{3}}=1.625\,mm\]You need to login to perform this action.
You will be redirected in
3 sec