Q122 / 180 · NEET 2019 (Odisha) · Physics
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.
- A$5 \frac{\lambda}{2}$
- B$10 \frac{\lambda}{2}$
- Correct answerC$9 \frac{\lambda}{2}$
- D$11 \frac{\lambda}{2}$
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, $\mathrm{n}=5$
$$
\therefore \quad \Delta y=[2(5)-1] \frac{\lambda}{2}=\frac{9 \lambda}{2}
$$
Hence, the correct answer is option: 'C'.