Question

In a bi-prism experiment fifth dark fringes are obtained at a point. If a thin transparent film is placed in the path of one of waves, then seventh bright fringe is obtained at the same point. The thickness of the film in terms of wavelength $X$ and refractive index $\mu$ will be

• A. $\frac{1.5\lambda }{(\mu -1)}$
• B. $1.5(\mu -1)\lambda$
• C. $2.5(\mu -1)\lambda$
• D. $\frac{2.5\lambda }{(\mu -1)}$

Answer 1

The correct option is D $\frac{2.5\lambda }{(\mu -1)}$

For 5th dark fringe to be obtained at the point, the path difference of the beams at that point =$\frac{9\lambda }{2}$.

For 7th bright fringe to be obtained at the same point, the new path difference of the beams at that point =$7\lambda$.

Due to insertion of the film of thickness $t$, path difference added to the beam=$t(\mu -1)$

Hence, $\frac{9\lambda }{2}+t(\mu -1)=7\lambda$

Hence, $t=\frac{2.5\lambda }{(\mu -1)}$