Respuesta :
Answer:
Fringe spacing if light is changed then [tex]\Delta y_2= 1.2 mm[/tex]
Explanation:
Given data is:
[tex]\lambda_{1}=600\,nm\\\\Distance \,\,between\,\, two \,\,bright\,\, fringes=\Delta\,y_{1}=1.8\,mm\\\\\lambda_{2}=400\nm\\\\d_{1}=d_{2}\\\\L_{1}=L_{2}\\\\\Delta\,y_{2}=?\\[/tex]
For Double slit experiment distance b/w two bright fringes is given as:
[tex]\Delta\,y=\frac{\lambda L}{d}\\\\For\,\, \lambda_{1}\\\\\Delta\,y_{1}=\frac{\lambda_{1} L_{1}}{d_{1}}\\\\\frac{L_1}{d_1}=\frac{\Delta y_1}{\lambda_1}--(1)\\\\For\,\,\lambda_2\\\\\Delta\,y_{2}=\frac{\lambda_{2} L_{2}}{d_{2}}\\\\Given \,\,that \\\\d_1=d_2,\,\,L_1=L_2\\\\so\\\\\Delta\,y_{2}=\frac{\lambda_{2} L_{1}}{d_{1}}\\\\Using\,\, (1) \,\,in \,\,above\\\\\Delta\,y_{2}=\frac{\lambda_{2} \Delta y_{1}}{\lambda_{1}}\\\\Substituting\,\, values\\\\[/tex][tex]\Delta y_2=\frac{(400nm)(1.8mm)}{(600nm)}\\\\\Delta y_2= 1.2\,mm[/tex]
The fringe spacing be if the light is changed to a wavelength of 400nm [tex]\Delta y_2=1.2mm[/tex]
What will be the fringe spacing be if the light is changed to a wavelength of 400nm?
It is given that
[tex]\lambda_1=600mm[/tex]
The distance between the two fringes will be [tex]\Delta Y_1=1.8mm[/tex]
[tex]\lambda _2=400mm[/tex]
[tex]d_1=d_2[/tex]
[tex]L_1=L_2[/tex]
[tex]\Delta Y_2=?[/tex]
As we know that the distance between fringes is given as
[tex]\Delta Y= \dfrac{\lambda L}{d}[/tex]
Now for [tex]\lambda _1[/tex]
[tex]\Delta Y_1=\dfrac{\lambda_1 L_1}{d_1}[/tex].............................(1)
Similarly now for [tex]\lambda _2[/tex]
[tex]\Delta Y_2=\dfrac{\lambda_2L_2}{d_2}[/tex]...............................(2)
From both the equations
[tex]\Delta Y_2=\dfrac{\lambda_2L_1}{d_1}[/tex]
[tex]\Delta Y_2=\dfrac{\lambda _2 \Delta Y_1}{\lambda _1}[/tex]
[tex]\Delta Y_2=\dfrac{(400\times1.8)}{600}[/tex]
[tex]\Delta Y_2=1.2mm[/tex]
Thus the fringe spacing be if the light is changed to a wavelength of 400nm [tex]\Delta y_2=1.2mm[/tex]
To know more about the Double slit experiment follow
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