In a Fraunhofer diffraction at a single slit | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For a single slit diffraction,the condition for minima is $d \sin 	heta = n\lambda$ where $n = 1, 2, 3, ...$For the first minimum $(n=1)$,$d \sin

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) For a single slit diffraction,the condition for minima is $d \sin 	heta = n\lambda$ where $n = 1, 2, 3, ...$For the first minimum $(n=1)$,$d \sin 30^{\circ} = \lambda$.Since $\sin 30^{\circ} = 0.5$,we have $d(0.5) = \lambda$,which implies $d = 2\lambda$.The condition for secondary maxima is $d \sin 	heta = (n + \frac{1}{2})\lambda$ where $n = 1, 2, 3, ...$For the first secondary maximum $(n=1)$,$d \sin 	heta = (1 + \frac{1}{2})\lambda = \frac{3}{2}\lambda$.Substituting $d = 2\lambda$ into the equation:$2\lambda \sin 	heta = \frac{3}{2}\lambda$$\sin 	heta = \frac{3}{4}$$	heta = \sin ^{-1}\left(\frac{3}{4}ight)$.

In a Fraunhofer diffraction at a single slit of width $d$ and incident light of wavelength $5500 \text{ Å}$,the first minimum is observed at an angle $30^{\circ}$. The first secondary maxima are observed at an angle $\theta=$

Similar Questions

$A$ slit of size $0.15 \ cm$ is placed at $2.1 \ m$ from a screen. On illuminating it by a light of wavelength $5 \times 10^{-5} \ cm$,the width of the central maxima will be:

The condition for observing Fraunhofer diffraction from a single slit is that the light wavefront incident on the slit should be

In Fraunhofer diffraction due to a single slit,find the path difference for the $n^{th}$ order minimum.

In a single slit diffraction pattern, the wavelength of light used is $628 \text{ nm}$ and slit width is $0.2 \text{ mm}$. The angular width of the central maximum is $\alpha \times 10^{-2} \text{ degrees}$. The value of $\alpha$ is . . . . . . . (Take $\pi = 3.14$)

$A$ screen is placed at $50 \ cm$ from a single slit,which is illuminated with light of wavelength $600 \ nm$. If the separation between the $1^{st}$ and $3^{rd}$ minima in the diffraction pattern is $3 \ mm$,then the slit width is: (in $mm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module