Question

Two narrow, parallel slits separated by 0.850 mm are illuminated by 600-nm light, and the viewing screen is 2.80 m away from the slits. (a) What is the phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe? (b) What is the ratio of the intensity at this point to the intensity at the center of a bright fringe?

Answer 1

Answer:

a) 7.947 radians

b) $\mathbf{\frac{I}{I_{max}}=0.4535}$

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

$tan\theta =\frac{y}{L}\\\Rightarrow tan\theta =\frac{2.5}{2800}=0.000892\\\Rightarrow \theta=tan^{-1}0.000892=0.05115^{\circ}$

$\Delta r= dsin\theta\Rightarrow \Delta r=dsin0.05115=7.589\times10^{-7}$

a) Phase difference

$\phi=\frac{2\pi}{\lambda}\Delta r\\\Rightarrow \phi=\frac{2\pi}{600\times 10^{-9}}7.589\times10^{-7}=7.947\ rad$

∴ 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) $\frac{I}{I_{max}}=cos^2\frac{\phi}{2}\\\Rightarrow \frac{I}{I_{max}}=cos^2\frac{7.947}{2}\\\Rightarrow \mathbf{\frac{I}{I_{max}}=0.4535}$

∴ Ratio of the intensity at this point to the intensity at the center of a bright fringe $\mathbf{\frac{I}{I_{max}}=0.4535}$