In a Young’s double slit experiment the slit is illuminated by a source having two wavelengths of 400 nm and 600 nm. If distance between slits, d = 1mm, and distance between the plane of the slit and screen, D = 10 m then the smallest distance from the central maximum where there is complete darkness is :
Text Solution
Verified by ExpertsD
Let us say, n th minima of 400 nm coincides with m th minima of 600 nm.
⇒ 
= = 
. 
⇒ 400 n = 600m + 100.
⇒ n = 
= (some integer or non-integer) + 0.25
Hence n can never be an integer. So no minima of 600 nm coincides with any minima of 400 nm.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems