In Young's double slit experiment,monochromat | vedclass

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

Question

(D) Let the intensity of light on the screen due to each slit be $I_0$.The maximum intensity at the centre of the screen is $I_{max} = 4I_0$.The inten

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(D) Let the intensity of light on the screen due to each slit be $I_0$.The maximum intensity at the centre of the screen is $I_{max} = 4I_0$.The intensity $I$ at a point where the phase difference is $\phi$ is given by $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 4I_0 \cos^2(\phi/2)$.We are given that the intensity becomes half of the maximum intensity,so $I = \frac{I_{max}}{2} = 2I_0$.Thus,$2I_0 = 4I_0 \cos^2(\phi/2) \implies \cos^2(\phi/2) = 1/2 \implies \cos(\phi/2) = 1/\sqrt{2}$.This gives $\phi/2 = \pi/4$,so the phase difference $\phi = \pi/2$.The phase difference is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.So,$\frac{2\pi}{\lambda} \Delta x = \frac{\pi}{2} \implies \Delta x = \frac{\lambda}{4}$.For small angles,the path difference $\Delta x = d \sin 	heta \approx d 	an 	heta = d(y/D)$.Equating the two,we get $d(y/D) = \lambda/4 \implies y = \frac{\lambda D}{4d}$.Given $\lambda = 5000 \ \mathring{A} = 5 	imes 10^{-7} \ m$,$d = 1.0 \ mm = 10^{-3} \ m$,and $D = 1.0 \ m$.$y = \frac{5 	imes 10^{-7} 	imes 1}{4 	imes 10^{-3}} = 1.25 	imes 10^{-4} \ m = 125 	imes 10^{-6} \ m$.Therefore,the distance is $125$.

Similar Questions

Two slits in Young's experiment have widths in the ratio $1 : 25$. The ratio of intensity at the maxima and minima in the interference pattern,$\frac{I_{max}}{I_{min}}$ is

In Young's double-slit experiment,the ratio of intensities of bright and dark fringes is $9$. What is the ratio of the intensities of the sources?

In a standard $YDSE$ setup,two coherent sources of light of intensity ratio $\beta$ produce an interference pattern. The value of $\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$ is equal to (where $I_{\max}$ and $I_{\min}$ are the maximum and minimum intensities of the resultant wave).

Two beams of monochromatic light with intensities $64 \ mW$ and $4 \ mW$ interfere constructively to produce an intensity of $100 \ mW$. If one of the beams is shifted by a phase angle $\phi$,the intensity is reduced to $84 \ mW$. The magnitude of $\phi$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module