Respuesta :
Answer:
1.21 cm.
Explanation:
Given that the wavelength of light is 550 nm.
And the distance between the slits is 0.10 mm.
And the distance to the screen is 2.2 m.
As we know that the lateral separation can be defined by the formula.
[tex]LS=\dfrac{m\lambda D}{d}[/tex]
Here, m is the order of maxima, [tex]\lambda[/tex] is the wavelength , d is the distance between the slits and D is the distance to the screen.
For first maxima m=1, and for second maxima m=2.
Difference in the lateral maxima will be,
[tex]\Delta LS=\dfrac{2\lambda D}{d}-\dfrac{1\lambda D}{d}\\\Delta LS=\dfrac{\lambda D}{d}[/tex]
Substitute all the variables in the above equation.
[tex]\Delta LS=\dfrac{550 nm\times 2.2m}{0.10mm}\\\Delta LS=\dfrac{550\times 10^{-9} nm\times 2.2m}{0.10\times 10^{-3} }\\ \Delta LS=121\times 10^{-4}m\\ \Delta LS=1.21 cm[/tex]
Therefore, the lateral separation difference between between first order and second order maxima is 1.21 cm.
