Optics Question 188
Question: The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young’s double-slit experiment is
[AIEEE 2004]
Options:
A) Infinite
B) Five
C) Three
D) Zero
Show Answer
Answer:
Correct Answer: B
Solution:
For maxima $ \Delta =d\sin \theta =n\lambda $
Therefore $ 2\lambda \sin \theta =n\lambda $
Therefore $ \sin \theta =\frac{n}{2} $ since value of sin q can not be greater 1. \ n = 0, 1, 2
Therefore only five maximas can be obtained on both side of the screen.
BETA
