Question

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

• A. 54
• B. 64
• C. 74
• D. 84

Answer 1

Answer: (D)

84

Answer 2

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 $n_2 = n_1 \big(\frac{\lambda_1}{\lambda_2}\big)$ $= 62 \times \frac{5893}{4358} = 84$