In a diffraction experiment from a single sli | vedclass

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

Question

(B) The linear width of the central maximum in a single-slit diffraction pattern is given by $w = \frac{2 \lambda D}{d}$,where $\lambda$ is the wavele

Vedclass · Accepted Answer

distance of the slit from the screen

Vedclass · Answer

(B) The linear width of the central maximum in a single-slit diffraction pattern is given by $w = \frac{2 \lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance from the slit to the screen,and $d$ is the slit width.The angular width of the central maximum is defined as the angle subtended by the central maximum at the slit,given by $	heta = \frac{w}{D} = \frac{2 \lambda}{d}$.From this formula,it is clear that the angular width depends only on the wavelength $\lambda$ and the slit width $d$. It does not depend on the distance $D$ between the slit and the screen.

In a diffraction experiment from a single slit,the angular width of the central maxima does $NOT$ depend upon

Similar Questions

In an optical fibre,the core and cladding are made with materials of refractive indices $1.5$ and $1.414$ respectively. To observe total internal reflection,what will be the range of the incident angle with the axis of the optical fibre?

Molten sodium chloride conducts electricity due to the presence of

Resistance of a decimolar solution between two electrodes $0.02 \ m$ apart and $0.0004 \ m^2$ in area was found to be $50 \ \Omega$. Specific conductance $(k)$ is : ........... $S \ m^{-1}$

If the line $ax + by + c = 0$ is a normal to the curve $xy = 1$,then

If $\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x=a \sin ^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c$ where $c$ is a constant of integration,then the ordered pair $(a, b)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module