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  • In Young's double slit experiment, the fringe width If entire arrangement is placed i

In Young's double slit experiment, the fringe width \mathrm{\beta.} If entire arrangement is placed in a liquid of refractive index \mathrm{n}, then the fringe width becomes:

Option: 1

\mathrm{\frac{\beta}{n+1}}


Option: 2

\mathrm{\mathrm{n} \beta}


Option: 3

\mathrm{\frac{\beta}{\mathrm{n}}}


Option: 4

\mathrm{\frac{\beta}{n-1}}


Answers (1)

best_answer

We have, \mathrm{\beta=\frac{\lambda D}{d}}

\mathrm{\because \quad}  The arrangement is placed in a liquid of refractive index n

\mathrm{ \therefore \quad \lambda^{\prime}=\frac{\lambda}{\mathrm{n}} }
Therefore fringe width
\mathrm{ \begin{aligned} \beta^{\prime} & =\frac{\lambda D}{\mathrm{dn}} \\ \text { or } \quad \beta^{\prime} & =\frac{\beta}{\mathrm{n}} \end{aligned} }

Posted by

Sanket Gandhi

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