(D) In Young's double slit experiment,the fringe width is given by $\beta = \frac{\lambda D}{d}$. Since the fringe width depends only on the wavelengt

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(D) In Young's double slit experiment,the fringe width is given by $\beta = \frac{\lambda D}{d}$. Since the fringe width depends only on the wavelength $\lambda$,the distance between slits $d$,and the distance to the screen $D$,the fringe width for dark and bright fringes is the same. Thus,the Assertion is incorrect.When white light is used as a source,the central fringe is white,and the subsequent fringes are coloured because different wavelengths have different fringe widths. Black fringes are not observed in the same way as monochromatic light. Thus,the Reason is also incorrect.

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

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Assertion: In Young's experiment,the fringe width for dark fringes is different from that for bright fringes.
Reason: In Young's double slit experiment,if the fringes are performed with a source of white light,then only black and bright fringes are observed.

AIIMS 2001 All AIIMS

A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
B
If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
C
If the Assertion is correct but Reason is incorrect.
D
If both the Assertion and Reason are incorrect.

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Wave Optics

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Young's Double Slit Experiment (YDSE)

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The graph shows the variation of fringe width $(X)$ versus the distance of the screen from the plane of the slits $(D)$ in Young's double-slit experiment (keeping other parameters constant,where $d$ is the distance between the slits). The wavelength of light used can be calculated as:

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$A$ beam of light consisting of two wavelengths, $650\; nm$ and $520\; nm$, is used to obtain interference fringes in a Young's double-slit experiment.
$(a)$ Find the distance of the third bright fringe on the screen from the central maximum for wavelength $650\; nm$.
$(b)$ What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

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In $Y.D.S.E.$,how many maxima can be obtained on the screen if the wavelength of light used is $200 \ nm$ and $d = 700 \ nm$?

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In a Young's double slit experiment,the source $S$ and the two slits $A$ and $B$ are vertical,with slit $A$ above slit $B$. The fringes are observed on a vertical screen $K$. The optical path length from $S$ to $B$ is increased very slightly (by introducing a transparent material of higher refractive index) and the optical path length from $S$ to $A$ is not changed. As a result,the fringe system on $K$ moves:

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In Young's double slit experiment,the central bright fringe can be identified

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Assertion: In Young's experiment,the fringe width for dark fringes is different from that for bright fringes.Reason: In Young's double slit experiment,if the fringes are performed with a source of white light,then only black and bright fringes are observed.

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The graph shows the variation of fringe width $(X)$ versus the distance of the screen from the plane of the slits $(D)$ in Young's double-slit experiment (keeping other parameters constant,where $d$ is the distance between the slits). The wavelength of light used can be calculated as:

In $Y.D.S.E.$,how many maxima can be obtained on the screen if the wavelength of light used is $200 \ nm$ and $d = 700 \ nm$?

In Young's double slit experiment,the central bright fringe can be identified

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Assertion: In Young's experiment,the fringe width for dark fringes is different from that for bright fringes.
Reason: In Young's double slit experiment,if the fringes are performed with a source of white light,then only black and bright fringes are observed.