question_answer

A YDSE is conducted in water\[({{\mu }_{1}})\] as shown in figure. A glass plate of thickness t and refractive index\[{{\mu }_{2}}\] is placed in the path of \[{{S}_{2}}\]. The optical path difference at O is

A) \[({{\mu }_{2}}-1)t\]

B) \[({{\mu }_{1}}-1)t\]

C) \[\left( \frac{{{\mu }_{2}}}{{{\mu }_{1}}}-1 \right)t\]

D) \[({{\mu }_{2}}-{{\mu }_{1}})t\]