(B) For the first diffraction minimum,the condition is given by $a \sin \theta = n\lambda$. For $n=1$,$a \sin \theta = \lambda$.Given $\theta = 30^{\c

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$\sin^{-1}\left(\frac{3}{4}\right)$

(B) For the first diffraction minimum,the condition is given by $a \sin \theta = n\lambda$. For $n=1$,$a \sin \theta = \lambda$.Given $\theta = 30^{\circ}$,we have $a = \frac{\lambda}{\sin 30^{\circ}} = 2\lambda$.For the first secondary maximum,the condition is $a \sin \theta' = \frac{3\lambda}{2}$.Substituting the value of $a = 2\lambda$ into the equation:$2\lambda \sin \theta' = \frac{3\lambda}{2}$$\sin \theta' = \frac{3\lambda}{2 \times 2\lambda} = \frac{3}{4}$.Therefore,$\theta' = \sin^{-1}\left(\frac{3}{4}\right)$.

Class 12 Physics Wave Optics Fresnel distance, Diffraction through circular slit, concept of Airy Disc, Rayleigh Criterion of Resolution , Fraunhofer diffraction

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In a Fraunhofer diffraction experiment at a single slit using light of wavelength $400 \ nm$,the first minimum is formed at an angle of $30^{\circ}$. The direction $\theta'$ of the first secondary maximum is given by

KCET 2008 All KCET

A
$\sin^{-1}\left(\frac{2}{3}\right)$
B
$\sin^{-1}\left(\frac{3}{4}\right)$
C
$\sin^{-1}\left(\frac{1}{4}\right)$
D
$\tan^{-1}\left(\frac{2}{3}\right)$

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Class 12

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Wave Optics

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Fresnel distance, Diffraction through circular slit, concept of Airy Disc, Rayleigh Criterion of Resolution , Fraunhofer diffraction

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In Fraunhofer diffraction pattern,slit width is $0.3 \ mm$ and screen is at $1.5 \ m$ away from the lens. If wavelength of light used is $4500 \ Å$,then the distance between the first minimum on either side of the central maximum is [ $\theta$ is small and measured in radian.] (in $mm$)

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In a Fraunhofer diffraction,light of wavelength $\lambda$ is incident on a slit of width $d$. The diffraction pattern is observed on a screen placed at a distance $D$. If the linear width of the central maximum is equal to two times the width of the slit,then the value of $D$ is:

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In a Fraunhofer diffraction experiment at a single slit using light of wavelength $400 \ nm$,the first minimum is formed at an angle of $30^{\circ}$. The direction $\theta'$ of the first secondary maximum is given by

Similar Questions

In Fraunhofer diffraction pattern,slit width is $0.3 \ mm$ and screen is at $1.5 \ m$ away from the lens. If wavelength of light used is $4500 \ Å$,then the distance between the first minimum on either side of the central maximum is [ $\theta$ is small and measured in radian.] (in $mm$)

In Fraunhofer diffraction,draw the graph of the intensity of light on the screen versus the diffraction angle $\theta$ and write its characteristics.

If the slit width is $2 \, mm$ and the wavelength of light used is $4000 \, Å$, then the Fresnel distance is nearly:

What do the Fraunhofer lines in the solar spectrum indicate?

In a Fraunhofer diffraction,light of wavelength $\lambda$ is incident on a slit of width $d$. The diffraction pattern is observed on a screen placed at a distance $D$. If the linear width of the central maximum is equal to two times the width of the slit,then the value of $D$ is:

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