Two waves of intensity ratio 1:9 cross each o | vedclass

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

Question

(B) Let the intensities of the two waves be $I_A = I_0$ and $I_B = 9I_0$.For incoherent waves,the resultant intensity is the sum of individual intensi

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) Let the intensities of the two waves be $I_A = I_0$ and $I_B = 9I_0$.For incoherent waves,the resultant intensity is the sum of individual intensities:$I_1 = I_A + I_B = I_0 + 9I_0 = 10I_0$.For coherent waves,the resultant intensity is given by $I_2 = I_A + I_B + 2\sqrt{I_A I_B} \cos \phi$,where $\phi = 60^{\circ}$.$I_2 = I_0 + 9I_0 + 2\sqrt{I_0 \cdot 9I_0} \cos 60^{\circ}$.$I_2 = 10I_0 + 2(3I_0) \cdot \frac{1}{2} = 10I_0 + 3I_0 = 13I_0$.Given $\frac{I_1}{I_2} = \frac{10}{x}$,we have $\frac{10I_0}{13I_0} = \frac{10}{x}$.Therefore,$x = 13$.

Two waves of intensity ratio $1:9$ cross each other at a point. The resultant intensities at the point are $I_1$ when the waves are incoherent and $I_2$ when the waves are coherent with a phase difference of $60^{\circ}$. If $\frac{I_1}{I_2} = \frac{10}{x}$,then $x = . . . . . . . . . . .$

Similar Questions

Four light sources produce the following four waves:
$(i)$ $y_1 = a \sin(\omega t + \phi_1)$
(ii) $y_2 = a \sin(2\omega t)$
(iii) $y_3 = d' \sin(\omega t + \phi_2)$
(iv) $y_4 = d' \sin(3\omega t + \phi)$
Superposition of which two waves gives rise to interference?

The colour of light is characterised by which of the following characteristics?

The frequency of a light wave with a wavelength of $500 \, \mathring A$ is $....... \, Hz$.

Coherent sources are those sources for which

The intensities of two coherent light waves are $I$ and $4I$. The maximum intensity of the resultant wave after interference is: (in $I$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module