(B) The angular width of the central maximum in a single-slit diffraction pattern is given by the formula $\theta = \frac{2\lambda}{a}$,where $\lambda

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decreases with increase of slit width.

(B) The angular width of the central maximum in a single-slit diffraction pattern is given by the formula $\theta = \frac{2\lambda}{a}$,where $\lambda$ is the wavelength of the light and $a$ is the width of the slit.From this relation,it is clear that the angular width $\theta$ is inversely proportional to the slit width $a$,i.e.,$\theta \propto \frac{1}{a}$.Therefore,as the slit width $a$ increases,the angular width of the central maximum decreases.

Class 12 Physics Wave Optics Single Slit Diffraction of Light

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$A$ parallel beam of monochromatic light falls normally on a single narrow slit. The angular width of the central maximum in the resulting diffraction pattern

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increases with increase of slit width.
B
decreases with increase of slit width.
C
decreases with decrease of slit width.
D
may increase or decrease with decrease of slit width.

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Single Slit Diffraction of Light

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Light of wavelength $\lambda$ is incident on a single slit of width $a$,and the distance between the slit and the screen is $D$. In the diffraction pattern,if the slit width is equal to the width of the central maximum,then $D=$

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$A$ narrow slit of width $2 \, mm$ is illuminated by monochromatic light of wavelength $500 \, nm$. The distance between the first minima on either side on a screen at a distance of $1 \, m$ is ........ $mm$.

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$A$ slit of width $a$ is illuminated by white light. For red light $(\lambda = 6500 \; \mathring{A})$,the first minima is obtained at $\theta = 30^\circ$. Then the value of $a$ will be

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The distance between the first and the sixth minima in the diffraction pattern of a single slit is $0.5 \, mm$. The screen is $0.5 \, m$ away from the slit. If the wavelength of light used is $5000 \, \mathring{A}$,then the slit width will be ....... $mm$.

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In the diffraction pattern of a single slit,the width of the secondary maxima compared to the central maximum is:

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$A$ parallel beam of monochromatic light falls normally on a single narrow slit. The angular width of the central maximum in the resulting diffraction pattern

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Light of wavelength $\lambda$ is incident on a single slit of width $a$,and the distance between the slit and the screen is $D$. In the diffraction pattern,if the slit width is equal to the width of the central maximum,then $D=$

$A$ narrow slit of width $2 \, mm$ is illuminated by monochromatic light of wavelength $500 \, nm$. The distance between the first minima on either side on a screen at a distance of $1 \, m$ is ........ $mm$.

$A$ slit of width $a$ is illuminated by white light. For red light $(\lambda = 6500 \; \mathring{A})$,the first minima is obtained at $\theta = 30^\circ$. Then the value of $a$ will be

The distance between the first and the sixth minima in the diffraction pattern of a single slit is $0.5 \, mm$. The screen is $0.5 \, m$ away from the slit. If the wavelength of light used is $5000 \, \mathring{A}$,then the slit width will be ....... $mm$.

In the diffraction pattern of a single slit,the width of the secondary maxima compared to the central maximum is:

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