Optics Question 1074
Question: The maximum intensity of fringes in Young’s experiment is I. If one of the slit is closed, then the intensity at that place becomes Io. Which of the following relation is true?
[NCERT 1982; MP PMT 1994, 99; BHU 1998; RPMT 1996; RPET 1999; AMU (Engg.) 1999]
Options:
A) I = Io
B) I = 2Io
C) I = 4Io
D) There is no relation between I and Io
Show Answer
Answer:
Correct Answer: C
Solution:
Suppose slit width’s are equal, so they produces waves of equal intensity say $ {I}’. $
Resultant intensity at any point $ I _{R}=4{I}’{{\cos }^{2}}\varphi $
where $ \varphi $ is the phase difference between the waves at the point of observation.
For maximum intensity $ \varphi =0^{o}\Rightarrow {I _{\max }}=4{I}’=I $ -(i)
If one of slit is closed,
Resultant intensity at the same point will be $ {I}’ $ only i.e. $ {I}’=I _{O} $ -(ii)
Comparing equation (i) and (ii) we get $ I=4I _{O} $
