menu search
brightness_auto
Feel free to answer or ask any questions.
If you are new here please see how to use or FAQ.
more_vert
Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source ($\lambda$ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :

(1) 2.87

(2) 2 nm

(3) 1.27 $\mu$m

(4) 2.05 $\mu$m
more_vert
Very lengthy question

1 Answer

more_vert
 
verified
Best answer

Ans. (3) 1.27 $\mu$m

Sol. $\Delta P = \mathrm{d}\sin\theta$ 

$= d\theta$

$={\mathrm{d}y\over D}={10^{-3}\times1.270mm\over1m}=1.27 \mu m$

thumb_up_off_alt 2 like thumb_down_off_alt 0 dislike
more_vert
Very short
Please explain more

Welcome to Jee Neet QnA, where you can ask questions and receive answers from other members of the community.


Join our Telegram group for live discussion.

Telegram Group

Subscribe our YouTube channel for video solutions with explanation.

YouTube Channel

Download Jee Neet QnA Books in PDF for offline learning.

Jee Neet QnA Books

1.2k questions

842 answers

383 comments

89 users

...