In a Young's double slit experiment with slit | vedclass

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

Question

(A) For bright fringes (maxima) in a Young's double slit experiment,the condition is $d \sin 	heta = n \lambda$,where $n$ is the order of the fringe.

Vedclass · Accepted Answer

$625\, nm, 500\, nm$

Vedclass · Answer

(A) For bright fringes (maxima) in a Young's double slit experiment,the condition is $d \sin 	heta = n \lambda$,where $n$ is the order of the fringe.Since $	heta$ is small,$\sin 	heta \approx 	heta = \frac{1}{40}\, rad$.Given $d = 0.1\, mm = 10^{-4}\, m$.The condition becomes $d 	heta = n \lambda$,so $\lambda = \frac{d 	heta}{n}$.Substituting the values: $\lambda = \frac{10^{-4} 	imes (1/40)}{n} = \frac{10^{-4}}{40n} = \frac{10^{-5}}{4n} = \frac{2500\, nm}{n}$.For $\lambda$ to be in the visible range ($380\, nm$ to $740\, nm$):If $n=4$,$\lambda = \frac{2500}{4} = 625\, nm$.If $n=5$,$\lambda = \frac{2500}{5} = 500\, nm$.Both $625\, nm$ and $500\, nm$ lie within the visible range. Thus,the wavelengths are $625\, nm$ and $500\, nm$.

Similar Questions

Two slits separated by $0.5 \,mm$ are illuminated by light of wavelength $500 \,nm$. The screen is at a distance of $120 \,cm$ from the slits. The phase difference between the interfering waves at a point $3 \,mm$ on the screen from the central bright fringe is ........... .

In Young's double slit experiment,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,the fringe width is $12 \ mm$. If the entire arrangement is placed in water of refractive index $\frac{4}{3}$,then the fringe width becomes (in $mm$):

In Young's double slit interference experiment,the slit separation is made $3$ folds. The fringe width becomes ......... times.

Vedclass Test Series

Exam Paper Generator

Online Exam Module