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.



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक