(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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Wave Optics

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Single Slit Diffraction of Light

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$A$ slit of size $0.15 \ cm$ is placed at $2.1 \ m$ from a screen. On illuminating it by a light of wavelength $5 \times 10^{-5} \ cm$,the width of the central maxima will be:

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In a single-slit diffraction pattern,the intensity of the central maximum is $I_0$. If the slit width is doubled,what will be the intensity of the central maximum?

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In a single slit diffraction experiment, the width of the slit is reduced. What happens to the linear width of the principal maxima?

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In a single slit diffraction pattern, the distance between the plane of the slit and the screen is $1.4 \text{ m}$. The width of the slit is $0.66 \text{ mm}$. The second maximum is formed at the distance of $2.8 \text{ mm}$ from the center of the screen. The wavelength of light used is (in $\text{ Å}$)

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In a single slit diffraction pattern, the wavelength of light used is $628 \text{ nm}$ and slit width is $0.2 \text{ mm}$. The angular width of the central maximum is $\alpha \times 10^{-2} \text{ degrees}$. The value of $\alpha$ is . . . . . . . (Take $\pi = 3.14$)

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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

Similar Questions

$A$ slit of size $0.15 \ cm$ is placed at $2.1 \ m$ from a screen. On illuminating it by a light of wavelength $5 \times 10^{-5} \ cm$,the width of the central maxima will be:

In a single-slit diffraction pattern,the intensity of the central maximum is $I_0$. If the slit width is doubled,what will be the intensity of the central maximum?

In a single slit diffraction experiment, the width of the slit is reduced. What happens to the linear width of the principal maxima?

In a single slit diffraction pattern, the wavelength of light used is $628 \text{ nm}$ and slit width is $0.2 \text{ mm}$. The angular width of the central maximum is $\alpha \times 10^{-2} \text{ degrees}$. The value of $\alpha$ is . . . . . . . (Take $\pi = 3.14$)

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