Respuesta :
Answer:
a) 7.947 radians
b) [tex]\mathbf{\frac{I}{I_{max}}=0.4535}[/tex]
Explanation:
y = Distance from central bright fringe = 2.5 mm
λ = Wavelength = 600 nm
L = Distance between screen and source = 2.8 m
d = Slit distance = 0.85 mm
[tex]tan\theta =\frac{y}{L}\\\Rightarrow tan\theta =\frac{2.5}{2800}=0.000892\\\Rightarrow \theta=tan^{-1}0.000892=0.05115^{\circ}[/tex]
[tex]\Delta r= dsin\theta\Rightarrow \Delta r=dsin0.05115=7.589\times10^{-7}[/tex]
a) Phase difference
[tex]\phi=\frac{2\pi}{\lambda}\Delta r\\\Rightarrow \phi=\frac{2\pi}{600\times 10^{-9}}7.589\times10^{-7}=7.947\ rad[/tex]
∴ Phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe is 7.947 radians
b) [tex]\frac{I}{I_{max}}=cos^2\frac{\phi}{2}\\\Rightarrow \frac{I}{I_{max}}=cos^2\frac{7.947}{2}\\\Rightarrow \mathbf{\frac{I}{I_{max}}=0.4535}[/tex]
∴ Ratio of the intensity at this point to the intensity at the center of a bright fringe [tex]\mathbf{\frac{I}{I_{max}}=0.4535}[/tex]
