Question
Question: Suppose that Young's double slit experiment is carried out with sodium (Yellow) light of wavelength ...
Suppose that Young's double slit experiment is carried out with sodium (Yellow) light of wavelength 589.3nm and the interference pattern is observed on a 100cm away. The 10th bright fringe has its centre at a distance of 12mm from the central maximum. Find out the separation between the slits.
A.0.49mmB.0.06mmC.0.7mmD.0.53mm
Solution
The fringe width can be found by taking the ratio of the product of order of the fringe, the wavelength of the light used and the distance between the slit and source to the distance between the slits. From this find out the distance between the slits by rearranging the equation. This will help you in answering this question.
Complete step by step answer:
It has been mentioned in the question that the order of the bright fringe formed will be,
n=10
The wavelength of the light used can be written as,
λ=589.3nm
Fringe width from the central maximum can be written as,
β=12mm
The distance between the slit and source can be written as,
D=100cm=1m
As we all know, the fringe width can be found using the equation,
β=dnλD
Let us substitute this in the equation. That is we can write that,
⇒12×10−3=d10×589.3×10−9×1
Rearrange this equation in terms of the distance between the sits. That is,