Two coherent sources P and Q produce interfer | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For constructive interference (bright bands),the path difference is $\Delta x = n\lambda$,where $n = 0, 1, 2, \dots$For destructive interference (

Vedclass · Accepted Answer

$2.7 	imes 10^{-4} 	ext{ cm}$

Vedclass · Answer

(B) For constructive interference (bright bands),the path difference is $\Delta x = n\lambda$,where $n = 0, 1, 2, \dots$For destructive interference (dark bands),the path difference is $\Delta x = (n + \frac{1}{2})\lambda$,where $n = 0, 1, 2, \dots$The $4^{	ext{th}}$ bright band corresponds to $n = 4$ $(\Delta x = 4\lambda)$ and the $5^{	ext{th}}$ bright band corresponds to $n = 5$ $(\Delta x = 5\lambda)$.$A$ dark band formed between the $4^{	ext{th}}$ and $5^{	ext{th}}$ bright band corresponds to $n = 4$ in the dark band formula: $\Delta x = (4 + \frac{1}{2})\lambda = 4.5\lambda$.Given $\lambda = 6000 	ext{ Å} = 6000 	imes 10^{-10} 	ext{ m} = 6 	imes 10^{-7} 	ext{ m} = 6 	imes 10^{-5} 	ext{ cm}$.Path difference $\Delta x = 4.5 	imes (6 	imes 10^{-5} 	ext{ cm}) = 27 	imes 10^{-5} 	ext{ cm} = 2.7 	imes 10^{-4} 	ext{ cm}$.

Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ bright band and the $5^{\text{th}}$ bright band. The wavelength of the light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is:

Similar Questions

Young's double-slit experiment is performed using green light,red light,and blue light,one at a time. The fringe widths recorded are $\beta_G, \beta_R,$ and $\beta_B$ respectively. Then:

In an interference pattern,the fringe width is $\beta$. If the frequency of the source is doubled,the fringe width becomes .....

Vedclass Test Series

Exam Paper Generator

Online Exam Module