Question

Two coherent beam of wavelength 5000 Å reaching a point would individually produce
intensities 1.44 and 4.00 units. If they reach there together, the intensity is 0.90 units.
Calculate the lowest phase difference with which the beam reach that point.​

Answer 1

Answer:give your phone number

Explanation: I will explain you

Answer 2

Answer:

The lowest phase difference is 161°.

Explanation:

Given that,

Wavelength = 5000 Å

Intensity I₁= 1.44 units

Intensity I₂=4.00 units

Intensity = 0.90 units

We need to calculate the phase difference

Using formula of intensity

$I=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos\delta$

Put the value into the formula

$0.9=1.44+4.00+2\sqrt{1.44\times4.00}\cos\delta$

$\cos\delta=\dfrac{0.90-(1.44)-(4.00)}{2\sqrt{1.44\times4.00}}$

$\cos\delta=-0.945$

$\cos(180-\delta)=0.945$

$180-\delta=\cos^{-1}(0.945)$

$-\delta=19.09-180$

$\delta=161^{\circ}$

Hence, The lowest phase difference is 161°.