JEE MAIN - Physics (2009 - No. 9)
A mixture of light, consisting of wavelength $$590$$ $$nm$$ and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the $$4$$th bright fringe of the unknown light. From this data, the wavelength of the unknown light is :
$$885.0$$ $$nm$$
$$442.5$$ $$nm$$
$$776.8$$ $$nm$$
$$393.4$$ $$nm$$
Explanation
Third bright fringe of known light coincides with the 4th bright fringe of the unknown light.
$$\therefore$$ $${{3\left( {590} \right)D} \over d} = {{4\lambda D} \over d}$$
$$ \Rightarrow \lambda = {3 \over 4} \times 590$$
$$ = 442.5\,nm$$
$$\therefore$$ $${{3\left( {590} \right)D} \over d} = {{4\lambda D} \over d}$$
$$ \Rightarrow \lambda = {3 \over 4} \times 590$$
$$ = 442.5\,nm$$