Respuesta :
Answer:
The fractional Intensity [tex]\frac{I}{I_{max}}[/tex] = 0.0146
Given:
wavelength of the light, [tex]\lambda = 546.1 nm = 546.1\times 10^{-9} m[/tex]
slit and screen separation difference, D = 130 cm = 1.3 m
distance of the point from the center of the principal maximum, y = 4.10 mm = 0.041 m
slit width, d = 0.420 mm = [tex]0.420\times 10^{-3}[/tex]
Solution:
To calculate the fractional intensity, we use the given formula:
[tex]\frac{I}{I_{max}} = \frac{sin^{2}\delta }{\frac({\delta}{2})^{2}}[/tex] (1)
[tex]\delta = \frac{\pi }{\lambda}dsin\theta [/tex]
For very small angle:
[tex]\delta = \frac{\pi dy}{\lambda D}[/tex] (2)
where
[tex]\delta = total phase angle [/tex]
[tex]\theta = angle of deviation[/tex]
Using eqn (2):
[tex]\delta = \frac{\pi \times 0.42\times 10^{-3}\times 4.1\times 10^{-3} }{546.1\times 10^{-9}\times 1.3} = 7.6202 radians[/tex]
Now, using eqn (1):
[tex]\frac{I}{I_{max}} = \frac{sin^{2}(7.6202) }{(\frac{7.6202}{2})^{2}} = 0.0146[/tex]
