Two beams of light having intensities I and 4 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The resultant intensity $I_R$ for two interfering beams with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by $I_R = I_1 + I_2

Vedclass · Accepted Answer

$4I$

Vedclass · Answer

(D) The resultant intensity $I_R$ for two interfering beams with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi)$.Given $I_1 = I$ and $I_2 = 4I$.At point $A$,the phase difference $\phi_A = \frac{\pi}{2}$.$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos\left(\frac{\pi}{2}\right) = 5I + 2(2I)(0) = 5I$.At point $B$,the phase difference $\phi_B = 2\pi$.$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(2\pi) = 5I + 2(2I)(1) = 5I + 4I = 9I$.The difference between the resultant intensities is $I_B - I_A = 9I - 5I = 4I$.

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi}{2}$ at point $A$ and $2\pi$ at point $B$. Find the difference between the resultant intensities at point $B$ and point $A$.

Similar Questions

In Young's double-slit experiment,the intensity at a point where the path difference is $\lambda$ is $k$. What will be the intensity at a point where the path difference is $\lambda/4$? ($\lambda$ = wavelength of light)

The ratio of maximum to minimum intensity due to the superposition of two waves is $\frac{49}{9}$. Then the ratio of the intensities of the component waves is:

If $\frac{I_1}{I_2} = \frac{16}{1},$ then $\frac{I_{\max}}{I_{\min}} = ?$

Obtain the formula for the intensity of light if the phase difference at a point from two coherent sources is $\phi$.

Two coherent sources of intensity ratio $\alpha$ interfere. The value of $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module