JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 12)
The figure shows a Young's double slit
experimental setup. It is observed that when a
thin transparent sheet of thickness t and
refractive index μ is put in front of one of the
slits, the central maximum gest shifted by a
distance equal to n fringe widths. If the
wavelength of light used is $$\lambda $$, t will be :
_9th_April_Morning_Slot_en_12_1.png)
$${{D\lambda } \over {a\left( {\mu - 1} \right)}}$$
$${{2nD\lambda } \over {a\left( {\mu - 1} \right)}}$$
$${{2D\lambda } \over {a\left( {\mu - 1} \right)}}$$
$${{n\lambda } \over {\left( {\mu - 1} \right)}}$$
Explanation
Path difference at central maxima $$\Delta $$x = ($$\mu $$ – 1)t, whole pattern will shift by same amount which will be given by
$$\left( {\mu - 1} \right)t{D \over d} = n{{\lambda D} \over d};$$ according to the question $$t = {{n\lambda } \over {\left( {\mu - 1} \right)}}$$
$$\left( {\mu - 1} \right)t{D \over d} = n{{\lambda D} \over d};$$ according to the question $$t = {{n\lambda } \over {\left( {\mu - 1} \right)}}$$
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