question_answer

A light of wavelength 630 nm incident on a pair of slits produces an interference pattern in which two successive bright fringes are separated by 7.2 mm.

Calculate the wavelength of another source of light which will produce interference fringes separated by 8.1 mm using same pair of slits.

Answer:

708.75 nm Given, \[{{\beta }_{1}}=7.2\times {{10}^{-3}}m,\]\[{{\beta }_{2}}=8.1\times {{10}^{-3}}m\] And      \[{{\lambda }_{1}}=630\times {{10}^{-9}}m\] \[\because \]       Fringe width, \[\beta =\frac{D\lambda }{d}\] Where, \[\lambda =\] wavelength, D = separation between slits and screen and d = separation between two slits. \[\Rightarrow \]   \[\frac{{{\beta }_{1}}}{{{\beta }_{2}}}=\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}\]  (\[\because \]D and d are same) Wavelength of another source of laser light \[\Rightarrow \]   \[{{\lambda }_{2}}=\frac{{{\beta }_{2}}}{{{\beta }_{1}}}\times {{\lambda }_{1}}=\frac{8.1\times {{10}^{-3}}}{7.2\times {{10}^{-3}}}\times 630\times {{10}^{-9}}m\] Or         \[{{\lambda }_{2}}=708.75\times {{10}^{-9}}m\] \[\therefore \]      \[{{\lambda}_{2}}=708.75\,\text{nm}\]