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Question: In Young’s double slit experiment, the slits are separated by \(0.5\ mm\) and the screen is placed \...

In Young’s double slit experiment, the slits are separated by 0.5 mm0.5\ mm and the screen is placed 1 m1\ m away from the slit. It is found that the 5th{{5}^{th}} bright fringes are at a distance of 4.13 mm4.13\ mm from the 2nd{{2}^{nd}} dark fringe. Find the wavelength of light used.

Explanation

Solution

Hint: In Young’s double slit experiment the position of bright fringes is given by the formula, y5th=(n2+1)(λDd){{y}_{{{5}^{th}}}}=\left( \dfrac{n}{2}+1 \right)\left( \dfrac{\lambda D}{d} \right). Now, using this formula we can easily find the wavelength of the light used in the experiment.

Formula used: y5th=(n2+1)(λDd){{y}_{{{5}^{th}}}}=\left( \dfrac{n}{2}+1 \right)\left( \dfrac{\lambda D}{d} \right)

Complete step by step answer:
In the question it is given that in Young’s double slit experiment, the slits are separated by 0.5 mm0.5\ mm and the screen is placed 1 m1\ m away from the slit. It is found that the fifth bright fringes are at a distance of 4.13 mm4.13\ mm from the second dark fringe and we are asked to find the wavelength of the light used. First of all, the nth bright fringe in young’s double slit experiment can be given by the formula,
y5th=(n2+1)(λDd){{y}_{{{5}^{th}}}}=\left( \dfrac{n}{2}+1 \right)\left( \dfrac{\lambda D}{d} \right)
Where, λ\lambda is wavelength of light, D is distance of screen from the slit, d is distance between two slits, and n is the number of fringes.
Now, in question it is given that,
d=0.5 mm=5×10−4 md=0.5\ mm=5\times {{10}^{-4}}\ m
D=1 mD=1\ m
Now, it is said that fifth bright is at 4.13 mm4.13\ mm distance from second dark which can be given mathematically as,
y5th=(n2+1)(λDd)=y2nd{{y}_{{{5}^{th}}}}=\left( \dfrac{n}{2}+1 \right)\left( \dfrac{\lambda D}{d} \right)={{y}_{{{2}^{nd}}}}
⇒y5th=(n2+1)(λDd)=4.13×10−3\Rightarrow {{y}_{{{5}^{th}}}}=\left( \dfrac{n}{2}+1 \right)\left( \dfrac{\lambda D}{d} \right)=4.13\times {{10}^{-3}}
Now, substituting the values given in the question in the expression we will get,
⇒(72)(λ(1)5×10−4)=4.13×10−3\Rightarrow \left( \dfrac{7}{2} \right)\left( \dfrac{\lambda \left( 1 \right)}{5\times {{10}^{-4}}} \right)=4.13\times {{10}^{-3}}
⇒7λ=4.13×10−3×5×10−4×2\Rightarrow 7\lambda =4.13\times {{10}^{-3}}\times 5\times {{10}^{-4}}\times 2
⇒λ=4.13×5×2×10−3×10−47\Rightarrow \lambda =\dfrac{4.13\times 5\times 2\times {{10}^{-3}}\times {{10}^{-4}}}{7}
⇒λ=5.9×10−7m=5.9×103×10−10mm\Rightarrow \lambda =5.9\times {{10}^{-7}}m=5.9\times {{10}^{3}}\times {{10}^{-10}}mm
⇒λ=5900A∘ \Rightarrow \lambda =5900\overset{\circ }{\mathop{\text{A}}}\,
Thus, wavelength of light is 5900A∘ 5900\overset{\circ }{\mathop{\text{A}}}\,.

Note: In this type of sums students might make mistakes in considering the fifth bright fringe distance equal to the second dark fringe and due to that the solution might become hard to solve for them. So, students should know this basic of young’s double slit experiment.