(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 the Fraunhofer diffraction experiment, $L$ is the distance between the screen and the obstacle, $b$ is the size of the obstacle, and $\lambda$ is the wavelength of the incident light. The general condition for the applicability of Fraunhofer diffraction is:

MediumAP EAMCET 2008

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Consider the following statements $A$ and $B$ and identify the correct answer:
$A$. Fresnel's diffraction pattern occurs when the source of light or the screen on which the diffraction pattern is seen, or when both are at a finite distance from the aperture.
$B$. Diffracted light can be used to estimate the helical structure of nucleic acids.

MediumTS EAMCET 2004

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What will be the angular width of the central maxima in Fraunhofer diffraction when light of wavelength $6000 \,\mathring{A}$ is used and the slit width is $12 \times 10^{-5} \, \text{cm}$? (in $rad$)

Medium

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$A$ single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light does the third secondary maximum in the diffraction pattern coincide with the second secondary maximum in the pattern for red light of wavelength $6500 Å$ (in $Å$)?

DifficultKCET 2007

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For what distance is ray optics a good approximation when the aperture is $3\; mm$ wide and the wavelength is $500\; nm$?

Easy

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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 the Fraunhofer diffraction experiment, $L$ is the distance between the screen and the obstacle, $b$ is the size of the obstacle, and $\lambda$ is the wavelength of the incident light. The general condition for the applicability of Fraunhofer diffraction is:

What will be the angular width of the central maxima in Fraunhofer diffraction when light of wavelength $6000 \,\mathring{A}$ is used and the slit width is $12 \times 10^{-5} \, \text{cm}$? (in $rad$)

$A$ single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light does the third secondary maximum in the diffraction pattern coincide with the second secondary maximum in the pattern for red light of wavelength $6500 Å$ (in $Å$)?

For what distance is ray optics a good approximation when the aperture is $3\; mm$ wide and the wavelength is $500\; nm$?

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