Assertion: In Young's double slit experiment, | vedclass

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

Question

(B) Let the two slits be $S_1$ and $S_2$. The point $P$ is directly opposite to slit $S_1$. Thus,$S_1P = D$.The distance $S_2P = \sqrt{D^2 + d^2} = D(

Vedclass · Accepted Answer

If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

Vedclass · Answer

(B) Let the two slits be $S_1$ and $S_2$. The point $P$ is directly opposite to slit $S_1$. Thus,$S_1P = D$.The distance $S_2P = \sqrt{D^2 + d^2} = D(1 + \frac{d^2}{D^2})^{1/2}$.Using the binomial theorem for $d The path difference $\Delta x = S_2P - S_1P = (D + \frac{d^2}{2D}) - D = \frac{d^2}{2D}$.For a dark fringe,the path difference must be an odd multiple of $\lambda/2$. For the first dark fringe,$\Delta x = \frac{\lambda}{2}$.Equating the two,$\frac{d^2}{2D} = \frac{\lambda}{2}$,which gives $\lambda = \frac{d^2}{D}$.Thus,$\lambda \propto d^2$. The Assertion is correct.The Reason states that for a dark fringe,the intensity is zero,which is a correct statement,but it does not explain why the wavelength is proportional to $d^2$ in this specific geometric configuration. Therefore,the Reason is not the correct explanation of the Assertion.

Similar Questions

In a Young's double-slit experiment,the experiment is performed with blue light of wavelength $4360 \; \mathring{A}$ and green light of wavelength $5460 \; \mathring{A}$. If the distance of the $4^{th}$ bright fringe from the central maximum is $x$,then:

Electrons accelerated from rest by an electrostatic potential are collimated and sent through a Young's double slit experiment. The fringe width is $\omega$. If the accelerating potential is doubled,then the width is now close to ............. $\omega$.

In Young's double-slit experiment,monochromatic light is used to illuminate the two slits $A$ and $B$. Interference fringes are observed on a screen placed in front of the slits. If a thin glass plate is placed normally in the path of the beam coming from slit $A$,then:

In Young's double-slit experiment,if the distance between the slits is reduced to half and the distance between the slit and the screen is doubled,then the fringe width

In Young's double slit experiment,if the widths of the slits are in the ratio $4 : 9$,the ratio of the intensity at maxima to the intensity at minima will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module