Previous Year NEET Question- Wave Optics L-4

Question: In a Young’ 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) $10 \frac{\lambda}{2}$

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

D) $11 \frac{\lambda}{2}$

Answer: $9 \frac{\lambda}{2}$#

Sol:#

Given, there is no initial phase difference. $\therefore \quad$ Initial phase $=\delta=0$ Again, phase difference $=\frac{2 \pi}{\lambda} \times$ path difference

$$ \Rightarrow \delta^{\prime}=\frac{2 \pi}{\lambda} \times \Delta x \Rightarrow \Delta x=\frac{\lambda}{2 \pi} \times \delta^{\prime} $$

Now, for the fifth minima we will consider $n=4$ as initial phase difference is zero.

$\therefore \quad$ For fifth minimum, $\delta^{\prime}=(8+1) \pi=9 \pi$

$\therefore \quad$ Path difference, $\Delta x=\frac{\lambda}{2 \pi} \times 9 \pi=\frac{9 \lambda}{2}$