(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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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 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

The angular width of the central maxima in Fraunhofer diffraction for $\lambda = 6000 \; \mathring{A}$ is $\theta_{0}$. When the same slit is illuminated by another monochromatic light,the angular width decreases by $30 \%$. The wavelength of this light is ....... $\mathring{A}$.

$A$ light beam of wavelength $800 \,nm$ passes through a single slit and is projected on a screen kept at $5 \,m$ away from the slit. What should be the slit width for the ray optics approximation to be valid (in $\,mm$)?

In Fraunhofer diffraction pattern, slit width is $0.2 \, mm$ and screen is at $2 \, m$ away from the lens. If wavelength of light used is $5000 \, \mathring{A}$, then the distance between the first minimum on either side of the central maximum is (assume $\theta$ is small and measured in radians).

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