Light of wavelength 520, nm passing through a | vedclass

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

Question

(A) From the given interference pattern,the angular width of a fringe is the distance between two consecutive minima or maxima. From the figure,the fi

Vedclass · Accepted Answer

$2 	imes 10^{-2}\, mm$

Vedclass · Answer

(A) From the given interference pattern,the angular width of a fringe is the distance between two consecutive minima or maxima. From the figure,the first minimum is at $	heta = 0.75^{\circ}$. The angular width of a fringe is $\Delta	heta = 0.75^{\circ}$. For a double slit experiment,the condition for the first minimum is $d \sin	heta = \frac{\lambda}{2}$. Since $	heta$ is very small,$\sin	heta \approx 	heta$ (in radians). Thus,$d 	heta = \frac{\lambda}{2} \Rightarrow d = \frac{\lambda}{2	heta}$. Convert $	heta$ to radians: $	heta = 0.75^{\circ} = 0.75 	imes \frac{\pi}{180} 	ext{ rad} = \frac{3}{4} 	imes \frac{\pi}{180} = \frac{\pi}{240} 	ext{ rad}$. Given $\lambda = 520 	imes 10^{-9} 	ext{ m}$. Substituting the values: $d = \frac{520 	imes 10^{-9}}{2 	imes (\pi / 240)} = \frac{520 	imes 10^{-9} 	imes 120}{\pi} \approx \frac{62400 	imes 10^{-9}}{3.14} \approx 1.987 	imes 10^{-5} 	ext{ m}$. Converting to mm: $d \approx 1.987 	imes 10^{-2} 	ext{ mm} \approx 2 	imes 10^{-2} 	ext{ mm}$. Therefore,the correct option is $A$.

Light of wavelength $520\, nm$ passing through a double slit produces an interference pattern of relative intensity versus deflection angle $\theta$ as shown in the figure. The separation $d$ between the slits is

Similar Questions

Light of wavelength $6000 \,\mathring A$ is incident on two slits. The distance between the slits is $0.1 \, cm$ and the screen is placed at a distance of $1 \, m$ from them. What is the distance between two consecutive minima in $mm$?

The correct curve between fringe width $\beta$ and the distance between the slits $(d)$ is:

In a Young's double slit experiment,the ratio of the amplitude of light coming from the slits is $2:1$. The ratio of the maximum to minimum intensity in the interference pattern is:

In $YDSE$ setup,light of wavelength $640 \, nm$ is used with $d = 0.8 \, mm$ and $D = 1 \, m$. If intensity at central maximum is $I_0$ and its position is $y = 0$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module