Question

In Young's experiment, the third bright band for light of wavelength $ 6000\,\mathring{A} $ coincides with the fourth bright band for another source of light in the same arrangement. Then the wavelength of second source is

• A. $ 3600\,\mathring{A} $
• B. $ 4000\,\mathring{A} $
• C. $ 5000\,\mathring{A} $
• D. $ 4500\,\mathring{A} $
• E. $ 5500\,\mathring{A} $

Answer 1

Answer: (D)

$ x=\frac{mD{{\lambda }_{1}}}{d}=\frac{(m+1)D{{\lambda }_{2}}}{d} $
$ \Rightarrow $ $ 3\times 6000=4{{\lambda }_{2}} $
Or $ {{\lambda }_{2}}=\frac{3\times 6000}{4} $
$ =4500\mathring{A} $