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  4. In Young's experiment when sodium light of wavelength 5893 mathring A is used, then 62 fringes are seen in the field of view. Instead, if violet light of wavelength 4358 mathring A is used, then the number of fringes that will be seen in the field of view will be

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Q. In Young's experiment when sodium light of wavelength 5893 $\mathring {A}$ is used, then 62 fringes are seen in the field of view. Instead, if violet light of wavelength 4358 $\mathring {A}$ is used, then the number of fringes that will be seen in the field of view will be

MP PMTMP PMT 2006Wave Optics

Solution:

Given $\lambda_1 = 5893 \mathring {A} , n_1 = 62, \lambda = 4358 \mathring {A}$
As field of view in case of both the wavelengths is same,
hence
$ n_1 \beta_1 = n_2 \beta_2$
as $ n_1 \big(\frac{D \lambda_1}{d}\big) = n_2 \big(\frac{D \lambda_2}{d}\big)$
or $ n_2 = n_1 \big(\frac{\lambda_1}{\lambda_2}\big)$
$= 62 \times \frac{5893}{4358} = 84$