Question

Two coherent light waves of intensity 5 x 10-2 Wm-2 each super-impose
and produce the interference pattern on a screen. At a point where the
path difference between the waves is 2/6, 9 being wavelength of the wave,
find the
(a) phase difference between the waves.
(b) resultant intensity at the point.
(C) resultant intensity in terms of the intensity at the maximum.​

Answer 1

Answer:

Explanation:

A) pi/3

B) 0.15 w/m^2

C) 0.75*Max Intensity

Answer 2

(a). The phase difference between the waves is $\dfrac{\pi}{3}$

(b). The resultant intensity at the point is 0.199 Wm⁻²

(c). The resultant intensity at the maximum is 0.2 Wm⁻²

Explanation:

Given that,

Intensity of each wave $I=5\times10^{-2}\ W/m^2$

(a). We need to calculate the phase difference between the waves

Using formula of phase difference

$\phi=\dfrac{2\pi}{\lambda}\times n\lambda$

$\phi=2\pi\times\dfrac{1}{6}$

$\phi=\dfrac{\pi}{3}$

(b). We need to calculate the resultant intensity at the point

Using formula of intensity

$I_{max}=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos\phi$

Put the value into the formula

$I_{max}=5\times10^{-2}+5\times10^{-2}+2\sqrt{5\times10^{-2}\times5\times10^{-2}}\cos\dfrac{\pi}{3}$

$I_{max}=0.199\ W/m^2$

(c). We need to calculate the resultant intensity in terms of the intensity at the maximum.​

Using formula of intensity

$I_{max}=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos\phi$

Here, $I_{1}=I_{2}$

Put the value into the formula

$I_{max}=I+I+2\sqrt{I^2}\cos0$

$I_{max}=4I$

$I_{max}=4\times5\times10^{-2}$

$I_{max}=0.2\ W/m^2$

Hence, (a). The phase difference between the waves is $\dfrac{\pi}{3}$

(b). The resultant intensity at the point is 0.199 Wm⁻²

(c). The resultant intensity at the maximum is 0.2 Wm⁻²

Learn more :

Topic : intensity

https://brainly.in/question/4702596