question_answer

The two slits at a distance of 1 mm are illuminated by the light of wavelength \[\text{6}.5\times \text{l}{{\text{0}}^{-\text{7}}}\text{m}.\] the interference fringes are observed on a serene placed at a distance of 1 m. The distance between third dark fringe and fifth bright fringe will be

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,                                 \[d=1\,mm=1\times {{10}^{-3}}m\] \[\therefore \]  \[{{\beta }_{5}}-{{\beta }_{3}}=\frac{5\times 1\times 6.5\times {{10}^{-7}}}{2\times {{10}^{-3}}}\]                 \[{{\beta }_{5}}-{{\beta }_{4}}=1.625\times {{10}^{-3}}m\]                 \[{{\beta }_{5}}-{{\beta }_{3}}=1.625\,mm\]