Class 12PhysicsWave OpticsFresnel distance, Diffraction through circular slit, concept of Airy Disc, Rayleigh Criterion of Resolution , Fraunhofer diffraction
EasyIn a Fraunhofer diffraction pattern,the slit width is $0.5 \ mm$ and the screen is at $2 \ m$ away from the lens. If the wavelength of the light used is $5500 \ Å$,then the distance between the first minima on either side of the central maximum is (assume $\theta$ is small and measured in radians). (in $mm$)
- A$1.1$
- B$2.2$
- C$4.4$
- D$5.5$
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Similar Questions
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).
$A$ Fraunhofer diffraction pattern due to a narrow slit is obtained on a screen placed at a distance $D$ from the slit whose slit width is $a$. The distance of the first secondary maximum from the central maximum is
For what distance is ray optics a good approximation when the aperture is $4 \, mm$ wide and the wavelength is $500 \, nm$ (in $, m$)?
Medium
View SolutionThe cause of Fraunhofer lines is
Easy
View SolutionIn 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:
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