question_answer

In a Young's double-slit experiment, the intensity at a point where the path difference is\[=\frac{T}{4}=\frac{2}{4}=0.5s\] (\[\pi /2\]being the wavelength of the light used), is \[{{v}_{a}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{({{10}^{-3}})}{OA}\]. If \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{({{10}^{-3}})}{\sqrt{{{(\sqrt{2})}^{2}}+{{(\sqrt{2})}^{2}}}}\] denotes the maximum intensity, \[{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{({{10}^{-3}})}{OB}\]is equal to[AIEEE 2007]

A) \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{({{10}^{-3}})}{2}\]

B) \[{{V}_{A}}-{{V}_{B}}=0\]

C) \[=\frac{\frac{1}{2}qV}{qV}=\frac{1}{2}\]

D) \[I={{I}_{0}}(1-{{e}^{-t/\tau }})\]

Correct Answer: D

Solution :

[d] As phase differencepath difference

i.e.,

As

or

or              (where,)