A parallel beam of light of wavelength 6000 t | vedclass

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

Question

(A) For a single slit diffraction, the condition for the $n^{th}$ minima is given by $a \sin 	heta = n \lambda$, where $a$ is the slit width, $\lambd

Vedclass · Accepted Answer

$2 	imes 10^{-3} 	ext{ rad}$

Vedclass · Answer

(A) For a single slit diffraction, the condition for the $n^{th}$ minima is given by $a \sin 	heta = n \lambda$, where $a$ is the slit width, $\lambda$ is the wavelength, and $	heta$ is the angular position.Given: $\lambda = 6000 	ext{ Å} = 6000 	imes 10^{-10} 	ext{ m} = 6 	imes 10^{-7} 	ext{ m}$ and $a = 0.3 	ext{ mm} = 0.3 	imes 10^{-3} 	ext{ m} = 3 	imes 10^{-4} 	ext{ m}$.For the first minima, $n = 1$.Substituting the values: $3 	imes 10^{-4} \sin 	heta = 1 	imes 6 	imes 10^{-7}$.$\sin 	heta = \frac{6 	imes 10^{-7}}{3 	imes 10^{-4}} = 2 	imes 10^{-3}$.Since $	heta$ is very small, $\sin 	heta \approx 	heta$.Therefore, $	heta = 2 	imes 10^{-3} 	ext{ rad}$.

$A$ parallel beam of light of wavelength $6000 \text{ Å}$ is diffracted by a single slit of width $0.3 \text{ mm}$. The angular position of the first minima of the diffracted light is

Similar Questions

Maximum diffraction takes place in a given slit for

To observe diffraction,the size of an obstacle:

$A$ 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:

$A$ slit of width $12 \times 10^{-5} \ cm$ is illuminated by monochromatic light of wavelength $6000 \ \mathring A$. Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern in degrees $(^o)$.

In a single slit diffraction pattern:

Vedclass Test Series

Exam Paper Generator

Online Exam Module