question_answer

In Fresnel's biprism experiment the width of 10 fringes is 2cm which are formed at a distance of 2 meter from the slit. If the wavelength of light is \[5100\overset{o}{\mathop{A}}\,\] then the distance between two coherent  sources will be

A) \[5.1\times {{10}^{-4}}m\]

B) \[5.1\times {{10}^{4}}m\]

C) \[5.1\times {{10}^{-4}}mm\]

D) \[10.1\times {{10}^{-4}}cm\]

Correct Answer: A

Solution :

[a] \[d=\frac{D\lambda }{\beta }\]      ...(i)                According to question, \[\lambda =5100\times {{10}^{-10}}m\] \[\beta =\frac{2}{10}\times {{10}^{-2}}m\]       ...(ii)   D=2m,  d=? From eqs. (i) and (ii) \[d=\frac{2\times 51\times {{10}^{-8}}}{2\times {{10}^{-3}}}=5.1\times {{10}^{-4}}m\]