Question

If one of the slits of a standard Young's double-slit experiment is covered by a thin parallel sides glass slab so that it transmits only one-half the light intensity of the other, then

• A. The fringe pattern will get shifted towards the covered slit
• B. The fringe pattern will get shifted away from the covered slit
• C. The bright fringes will become less bright and the dark ones become more bright
• D. The fringe width will remain unchanged

Answer 1

Answer: (A), (C), (D)

The fringe width will remain unchanged

Due to the insertion of glass slab, the light through the slab would have a path difference integrated due to traveling in a denser medium. For central maxima to form, the path difference at the point should be zero. Hence the light from the slab must travel lesser than that from the uncovered slit so that overall both ray travel equivalent optical path. Hence the pattern shifts towards the covered slit.

The total intensity at each point of maxima is given by $I_{max}=(\sqrt{I}+\sqrt{I_2}{)}^2={(\sqrt{I}+\sqrt{\frac{I}{2}})}^2<4I_0$

Hence, bright fringes are less bright.

Similarly total intensity of minima is
$I_{min}={(\sqrt{I}-\sqrt{\frac{I}{2}})}^2>0$
But previously it was zero. Hence the dark fringes are more bright.

Since $\beta =\frac{D\lambda }{d}$ fringe width does not change.

So, options A, C, and D are correct.