In a Young's double slit experiment, if there is no initial...

In a Young's double slit experiment, if there is no initial phase-difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference.

Options

A $5 \frac{\lambda}{2}$
B $10 \frac{\lambda}{2}$
C $9 \frac{\lambda}{2}$
D $11 \frac{\lambda}{2}$

Correct answer

C. $9 \frac{\lambda}{2}$

Step-by-step solution

In a YDSE, the path difference for

\mathrm{n}

th minima is given by

\Delta

y=(2 n-1)

\frac{\lambda}{2}

For 5 th minima,

\mathrm{n}=5

\therefore \quad

\Delta

y=[2(5)-1]

\frac{\lambda}{2}=\frac{9 \lambda}{2}

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