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In the figure shown in a $YSDE$ , a parallel beam of light is incident on the slits from a medium of refractive index $n_{1}$ . The wavelength of light in this medium is $\lambda _{1}$ . A transparent slab of thickness ' $t$ ' and refractive index $n_{3}$ is put infront of one slit. The medium between the screen and the plane of the slits is $n_{2}$ Find the phase difference between the light waves reaching point ' $O$ ' (symmetrical, relative to the slits)

Option: 1
$\frac{2\pi }{n_{1}\lambda _{1}}\left ( n_{3}-n_{2} \right )t$

Option: 2
$\frac{2\pi }{\lambda _{1}}\left ( n_{3}-n_{2} \right )t$

Option: 3
$\frac{2\pi n_{1}}{n_{2}\lambda _{1}}\left ( \frac{n_{3}}{n_{2}}-1 \right )t$

Option: 4
$\frac{2\pi n_{1}}{\lambda _{1}}\left ( n_{3}-n_{2} \right )t$

Answers (1)

best_answer

Relation between phase & path difference -

$\Delta \phi = \frac{2\pi }{\lambda }\times \Delta x$

- wherein

$\Delta \phi =$ Phase difference

$\Delta x=$ Path Difference

$\lambda =$ Wavelength

light wavelength in medium $n_{1}$ is = $\lambda _{1}$

$\Rightarrow$ Wavelength in vacuum = $\lambda _{0}$ = $n_{1}\lambda _{1}$

The path difference between the lights waves reaching point 0 $=\left ( n_{3} -n_{2}\right )t$

=extra path which the light from $S_{1}$ travelled compared to the path from $S_{2}$ Corresponding phase difference

$=\frac{2\pi }{\lambda _{0}}\ (path \ difference)$ = $\frac{2\pi }{n_{1}\lambda _{1}}\left ( n_{3} -n_{2}\right )t$

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