Optics Question 1138
Question: In an interference experiment, third bright fringe is obtained at a point on the screen with a light of 700 nm. What should be the wavelength of the light source in order to obtain 5th bright fringe at the same point
[KCET 2003]
Options:
A) 500 nm
B) 630 nm
C) 750 nm
D) 420 nm
Show Answer
Answer:
Correct Answer: D
Solution:
$ n _{1}{\lambda _{1}}=n _{2}{\lambda _{2}}\Rightarrow 3\times 700=5\times {\lambda _{2}}\Rightarrow {\lambda _{2}}=420\ nm $
BETA
