(A) The distance between two consecutive dark fringes is known as the fringe width,denoted by $\beta$.The formula for fringe width is $\beta = \frac{\

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(A) The distance between two consecutive dark fringes is known as the fringe width,denoted by $\beta$.The formula for fringe width is $\beta = \frac{\lambda D}{d}$.Given:Wavelength $\lambda = 5000 \ \mathring{A} = 5000 \times 10^{-10} \ m = 5 \times 10^{-7} \ m$.Distance between screen and slit $D = 1 \ m$.Slit separation $d = 0.2 \ cm = 0.2 \times 10^{-2} \ m = 2 \times 10^{-3} \ m$.Substituting these values into the formula:$\beta = \frac{5 \times 10^{-7} \times 1}{2 \times 10^{-3}} \ m$$\beta = 2.5 \times 10^{-4} \ m$To convert this into $mm$,multiply by $10^3$:$\beta = 2.5 \times 10^{-4} \times 10^3 \ mm = 0.25 \ mm$.

Class 12 Physics Wave Optics Young's Double Slit Experiment (YDSE)

Easy

In a Young's double slit experiment,the slit separation is $0.2 \ cm$,the distance between the screen and slit is $1 \ m$. The wavelength of the light used is $5000 \ \mathring{A}$. The distance between two consecutive dark fringes (in $mm$) is

A
$0.25$
B
$0.26$
C
$0.27$
D
$0.28$

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Wave Optics

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Young's Double Slit Experiment (YDSE)

In $\text{YDSE}$,$S_1$ and $S_2$ have the same intensity $I_0$. Column-$I$ shows the distance $x$ of a point $P$ from the central point $O$ on the screen,and Column-$II$ shows the intensity at $P$. Match Column-$I$ with Column-$II$. (Wavelength is $\lambda$)

Column-$I$ Column-$II$

$(A) x = \frac{D \lambda}{d}$ $(P) I_0$

$(B) x = \frac{D \lambda}{4d}$ $(Q) 2 I_0$

$(C) x = \frac{D \lambda}{3d}$ $(R) 3 I_0$

$(D) x = \frac{D \lambda}{6d}$ $(S) 4 I_0$

Column-$I$	Column-$II$
$(A) x = \frac{D \lambda}{d}$	$(P) I_0$
$(B) x = \frac{D \lambda}{4d}$	$(Q) 2 I_0$
$(C) x = \frac{D \lambda}{3d}$	$(R) 3 I_0$
$(D) x = \frac{D \lambda}{6d}$	$(S) 4 I_0$

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In Young's double-slit experiment,both slits are illuminated by a laser beam and the interference pattern was observed on a screen. If the viewing screen is moved farther from the slit,what happens to the interference pattern?

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$A$ beam of monoenergetic electrons,which have been accelerated from rest by a potential $U$,is used to form an interference pattern in a Young's double slit experiment. The electrons are now accelerated by potential $4U$. Then,the fringe width

KVPY 2015

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In Young's double slit experiment,we get $60$ fringes in the field of view using monochromatic light of wavelength $4000 \; \mathring{A}$. If we use monochromatic light of wavelength $6000 \; \mathring{A}$,then the number of fringes obtained in the same field of view is

Easy

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In Young's double-slit experiment,the fringe width on the screen is $0.2 \, mm$. If the wavelength of the light used is increased by $10\%$ and the distance between the two slits $S_1$ and $S_2$ is also increased by $10\%$,the new fringe width will be ....... $mm$.

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In a Young's double slit experiment,the slit separation is $0.2 \ cm$,the distance between the screen and slit is $1 \ m$. The wavelength of the light used is $5000 \ \mathring{A}$. The distance between two consecutive dark fringes (in $mm$) is

Similar Questions

In Young's double-slit experiment,both slits are illuminated by a laser beam and the interference pattern was observed on a screen. If the viewing screen is moved farther from the slit,what happens to the interference pattern?

$A$ beam of monoenergetic electrons,which have been accelerated from rest by a potential $U$,is used to form an interference pattern in a Young's double slit experiment. The electrons are now accelerated by potential $4U$. Then,the fringe width

In Young's double slit experiment,we get $60$ fringes in the field of view using monochromatic light of wavelength $4000 \; \mathring{A}$. If we use monochromatic light of wavelength $6000 \; \mathring{A}$,then the number of fringes obtained in the same field of view is

In Young's double-slit experiment,the fringe width on the screen is $0.2 \, mm$. If the wavelength of the light used is increased by $10\%$ and the distance between the two slits $S_1$ and $S_2$ is also increased by $10\%$,the new fringe width will be ....... $mm$.

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