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 \[{{\lambda }_{2}}=708.75\times {{10}^{-9}}m\] \[\therefore \] \[{{\lambda}_{2}}=708.75\,\text{nm}\]
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