In Young's double slit experiment, we get 15 fringes per cm on the screen, using light of wavelength 5600 Ã…. For the same setting, how many fringes per cm will be obtained with light of wavelength 7000 Ã…?
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In double-slit experiments, the number of fringes per unit length is inversely proportional to the wavelength.
The number of fringes per unit length \( N \) is inversely proportional to the wavelength \( \lambda \):
\[
N \propto \frac{1}{\lambda}
\]
Thus, the ratio of the number of fringes for the two wavelengths is:
\[
\frac{N_1}{N_2} = \frac{\lambda_2}{\lambda_1}
\]
Substituting the values:
\[
\frac{N_1}{15} = \frac{7000}{5600}
\]
Solving for \( N_1 \):
\[
N_1 = 12
\]
Thus, the number of fringes per cm is 12 for the wavelength 7000 Ã….