Question

 A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with \(10^{\text {th }}\) bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same \(10^{\text {th }}\) bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be _____ mm .

Answer 1

Answer is : 11

In case of YDSE the distance of \(\mathrm{n}^{\text {th }}\) maxima from central maxima is given by

\(\mathrm{Y}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) 

Here n, D \(\&\) d are same

So, \(\mathrm{y} \times \lambda\)  

\(\Rightarrow \frac{\mathrm{y}_{2}}{\mathrm{y}_{1}}=\frac{\lambda_{2}}{\lambda_{1}} \Rightarrow \frac{\mathrm{y}_{2}}{10 \mathrm{~mm}}=\frac{660 \mathrm{~nm}}{600 \mathrm{~nm}}\) 

\(\Rightarrow y_{2}=11 \mathrm{~mm}\)