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Question: In a single slit diffraction the distance between slit & screen is \(1\;{\rm{m}}\). The size of the ...

In a single slit diffraction the distance between slit & screen is 1  m1\;{\rm{m}}. The size of the slit is 0.7  mm0.7\;{\rm{mm}} & second maximum is formed at a distance of 2  mm2\;{\rm{mm}} from the center of the screen, then find out the wavelength of light.

Explanation

Solution

In the solution we will use the general equation of bright fringe for calculating the wavelength of light. They show the relation between the wavelength of light, number of fringes, size of slit and the angle of diffraction. We will also use some basic trigonometry for calculating the angle of diffraction.

Complete step by step solution:
Given:
The distance between slit and screen is, D=1  mD = 1\;{\rm{m}}.
The size of the slit, d is d=0.7  mm=0.7  mm×1  m1000  mm=7×10−4  md = 0.7\;{\rm{mm}} = 0.7\;{\rm{mm}} \times \dfrac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}} = 7 \times {10^{ - 4}}\;{\rm{m}}.
Distance between second maxima and center of screen is, y=2  mm=  2  mm×1  m1000  mm=2×10−3  my = 2\;{\rm{mm}} = \;2\;{\rm{mm}} \times \dfrac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}} = 2 \times {10^{ - 3}}\;{\rm{m}}.

The general equation for calculating the wavelength of n bright fringe is,
(n+12)λ=dsin⁡θ\left( {n + \dfrac{1}{2}} \right)\lambda = d\sin \theta
Here, λ\lambda is wavelength of light

For secondary maxima, substitute n=2 in the above expression,
(2+12)λ=dsin⁡θ 5λ2=dsin⁡θ λ=25dsin⁡θ\begin{array}{l} \left( {2 + \dfrac{1}{2}} \right)\lambda = d\sin \theta \\\ \dfrac{{5\lambda }}{2} = d\sin \theta \\\ \lambda = \dfrac{2}{5}d\sin \theta \end{array} … (i)
If θ\theta is small then sin⁡θ=tan⁡θ\sin \theta = \tan \theta .tan⁡θ\tan \theta is given by,
tan⁡θ=yD\tan \theta = \dfrac{y}{D}
If θ\theta is very small than tan⁡θ=θ\tan \theta = \theta and the expression becomes,
θ=yD\theta = \dfrac{y}{D}
Substitute in equation (i),
λ=2yd5D\lambda = \dfrac{{2yd}}{{5D}}
Substitute the necessary value and calculate wavelength,
λ=2(7×10−4  m)(2×10−3  m)5(1  m)\lambda = \dfrac{{2\left( {7 \times {{10}^{ - 4}}\;{\rm{m}}} \right)\left( {2 \times {{10}^{ - 3}}\;{\rm{m}}} \right)}}{{5\left( {1\;{\rm{m}}} \right)}}
λ=56×10−8  m\lambda = 56 \times {10^{ - 8\;}}{\rm{m}}
As, 1 Angstrom is equal to 10−10  m{10^{ - 10}}\;{\rm{m}}. Converting the wavelength in angstrom,
λ=56×10−8  m  =56×10−8  m×1  A010−10  m=5600  A0  \lambda {\rm{ = 56}} \times {\rm{1}}{{\rm{0}}^{ - 8}}\;{\rm{m}}\; = 56 \times {10^{ - 8}}\;{\rm{m}} \times \dfrac{{1\;\mathop {\rm{A}}\limits^{\rm{0}} }}{{{{10}^{ - 10}}\;{\rm{m}}}} = 5600\;\mathop {\rm{A}}\limits^{\rm{0}} \;
Therefore, the wavelength of the light used is 5600  A05600\;\mathop {\rm{A}}\limits^{\rm{0}} .

Note: Make sure to use the approximation that the angle of diffraction is very small. Convert the units very carefully. First convert sin⁡θ\sin \theta into tan⁡θ\tan \theta then use the basic formula of trigonometry. Also, remember that it is the case of maxima, so do not use the formula of minima.