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(D) Due to the insertion of a glass slab, the light passing through the slab experiences an additional optical path difference because it travels in a

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(D) Due to the insertion of a glass slab, the light passing through the slab experiences an additional optical path difference because it travels in a denser medium. For the central maxima to form, the net path difference at the point must be zero. Consequently, the light from the slab must travel a shorter geometric distance compared to the light from the uncovered slit to compensate for the optical path difference. This causes the entire fringe pattern to shift towards the covered slit.The intensity at any point in the interference pattern is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$. The intensity of maxima is $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$. Since one slit transmits only half the intensity of the other $(I_1 Similarly, the intensity of minima is $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$. Since $I_1 
eq I_2$, $I_{min} > 0$, meaning the dark fringes are no longer perfectly dark and become brighter.Finally, the fringe width is given by $\beta = \frac{D \lambda}{d}$. Since $D$, $\lambda$, and $d$ remain unchanged, the fringe width remains constant.

If one of the slits of a standard $YDSE$ apparatus is covered by a thin parallel-sided glass slab so that it transmits only one-half of the light intensity of the other, then:

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