Question
Question: If in Young’s double-slit experiment, the slit width is 3cm, the separation between slits and screen...
If in Young’s double-slit experiment, the slit width is 3cm, the separation between slits and screen is 70cm and wavelength of light is 1000A∘ , then fringe width will be (μ=1.5)
(A) 2.3×10−3cm
(B) 2.3×10−4m
(C) 2.3×10−5cm
(D) 2.3×10−6m
Solution
Use the equation to find the nth order bright fringe and (n+21)th order dark fringe. Subtract both the values for bright and dark fringe for the same value n . This subtracted value will give the fringe width in Young’s double slit experiment.
Complete step by step solution:
Let the fringe width be denoted by β , slit width be represented by d , the distance of separation of slit and screen be denoted by D , and let the wavelength of light be denoted by λ . Let n denote the number of fringe under consideration. Let y denote the position of bright or dark fringe.
The position of the nth bright fringe is given by
y=dnDλ
The position for dark fringe is given by
y=(n+21)dDλ
The difference between the two will give the fringe width:
β=dDλ
Substituting the given values of D ,λ and d , we get the fringe width as
β=3×10−270×10−2×1000×10−10
We have converted the units of distance to its SI unit.
Evaluating the above equation, we get
β=2.34×10−6m
Therefore, option (D) is the correct option.
Additional information:
Young’s double-slit experiment was first performed by Thomas Young in the year 1801 . The experiment proves the wave nature of light. It involves the superposition of light to form a constructive interference and a destructive interference fringe pattern. It also shows the probabilistic behavior of a wave in quantum mechanics.
Note:
The SI unit of length is the meter. To convert angstrom to meter, we multiply by 10−10 .
It can be noted from the equation to find the fringe width that the fringe width directly depends on the distance of separation between the slits and the screen. So, the farther the distance, the bigger will be the fringe width.