(D) In single slit diffraction,the condition for the $n^{th}$ dark band is given by $d \sin \theta = n \lambda$. For small angles,$\sin \theta \approx

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(D) In single slit diffraction,the condition for the $n^{th}$ dark band is given by $d \sin \theta = n \lambda$. For small angles,$\sin \theta \approx \tan \theta = \frac{y}{D}$,where $y$ is the distance from the central bright band. Substituting this into the condition,we get $d \left( \frac{y}{D} \right) = n \lambda$. Thus,the distance of the $n^{th}$ dark band from the central bright band is $y_n = \frac{n \lambda D}{d}$. For the second dark band,we set $n = 2$. Therefore,$y_2 = \frac{2 \lambda D}{d}$.

Class 12 Physics Wave Optics Single Slit Diffraction of Light

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$A$ parallel beam of light of wavelength $\lambda$ is incident normally on a single slit of width $d$. Diffraction bands are obtained on a screen placed at a distance $D$ from the slit. The second dark band from the central bright band will be at a distance given by

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A
$\lambda dD$
B
$\frac{\lambda D}{2d}$
C
$\frac{2\lambda d}{D}$
D
$\frac{2\lambda D}{d}$

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Class 12 Questions

Class 12

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Wave Optics

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Single Slit Diffraction of Light

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The angular width of the central maxima in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength $6000 \ \mathring A$. When the slit is illuminated by light of another wavelength,the angular width decreases by $30\%$. The wavelength of this light will be ...... $\mathring A$.

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$A$ single slit of width $0.1\, mm$ is illuminated by a parallel beam of light of wavelength $6000\, \mathring{A}$ and diffraction bands are observed on a screen $0.5\, m$ from the slit. The distance of the third dark band from the central bright band is ........ $mm$.

MediumJEE MAIN 2017

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$A$ screen is placed at a distance $50\,cm$ from a single slit,which is illuminated with light of wavelength $690\,nm$. If the distance between the first and third minima is $3.00\,mm$,what is the width of the slit (in $,mm$)?

Medium

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In a diffraction pattern due to a single slit,the angular width of the central maxima becomes half when the wavelength of the light used is changed from $\lambda$ to $7000 Å$. Then the value of $\lambda$ is (in $Å$)

EasyAP EAMCET 2017

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$A$ parallel beam of light of wavelength $500\; nm$ falls on a narrow slit and the resulting diffraction pattern is observed on a screen $1\;m$ away. It is observed that the first minimum is at a distance of $2.5\;mm$ from the centre of the screen. Find the width of the slit. (in $; mm$)

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$A$ parallel beam of light of wavelength $\lambda$ is incident normally on a single slit of width $d$. Diffraction bands are obtained on a screen placed at a distance $D$ from the slit. The second dark band from the central bright band will be at a distance given by

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$A$ single slit of width $0.1\, mm$ is illuminated by a parallel beam of light of wavelength $6000\, \mathring{A}$ and diffraction bands are observed on a screen $0.5\, m$ from the slit. The distance of the third dark band from the central bright band is ........ $mm$.

$A$ screen is placed at a distance $50\,cm$ from a single slit,which is illuminated with light of wavelength $690\,nm$. If the distance between the first and third minima is $3.00\,mm$,what is the width of the slit (in $,mm$)?

In a diffraction pattern due to a single slit,the angular width of the central maxima becomes half when the wavelength of the light used is changed from $\lambda$ to $7000 Å$. Then the value of $\lambda$ is (in $Å$)

$A$ parallel beam of light of wavelength $500\; nm$ falls on a narrow slit and the resulting diffraction pattern is observed on a screen $1\;m$ away. It is observed that the first minimum is at a distance of $2.5\;mm$ from the centre of the screen. Find the width of the slit. (in $; mm$)

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