In a single slit diffraction experiment,for a | vedclass

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

Question

(B) The half-angular width of the central maxima for a single slit of width $d$ is given by $	heta = \frac{\lambda}{d}$.For the first case: $	heta =

Vedclass · Accepted Answer

$q: 1$

Vedclass · Answer

(B) The half-angular width of the central maxima for a single slit of width $d$ is given by $	heta = \frac{\lambda}{d}$.For the first case: $	heta = \frac{\lambda}{d} \implies d = \frac{\lambda}{	heta}$.For the second case: $q	heta = \frac{p\lambda}{d'} \implies d' = \frac{p\lambda}{q	heta}$.The half-angular width of the $n^{th}$ secondary maxima is given by $	heta_n = \frac{(2n+1)\lambda}{2d}$.For the first secondary maxima $(n=1)$,$	heta_{s1} = \frac{3\lambda}{2d}$ and $	heta_{s2} = \frac{3(p\lambda)}{2d'}$.The ratio is $\frac{	heta_{s1}}{	heta_{s2}} = \frac{3\lambda / 2d}{3p\lambda / 2d'} = \frac{1}{d} \cdot \frac{d'}{p} = \frac{d'}{pd}$.Substituting $d' = \frac{p\lambda}{q	heta}$ and $d = \frac{\lambda}{	heta}$,we get $\frac{	heta_{s1}}{	heta_{s2}} = \frac{p\lambda / q	heta}{p(\lambda / 	heta)} = \frac{1}{q}$.Thus,the ratio is $1:q$,which corresponds to the inverse of the given options if interpreted as $q:1$ or $1:q$. Given the standard form,the ratio of widths is $1:q$.

Similar Questions

$A$ screen is placed $50\,cm$ from a single slit,which is illuminated with $6000\,\mathring{A}$ light. If the distance between the first and third minima in the diffraction pattern is $3\,mm,$ the width of the slit is ......$mm$.

The distance between the first and the sixth minima in the diffraction pattern of a single slit is $0.5 \, mm$. The screen is $0.5 \, m$ away from the slit. If the wavelength of light used is $5000 \, \mathring{A}$,then the slit width will be ....... $mm$.

$A$ slit of width $d$ is illuminated by white light. The first minimum for red light $(\lambda = 6500\,\mathring{A})$ will fall at $\theta = 30^\circ$ when $d$ will be

If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is $30^{\circ}$ in a single slit diffraction pattern recorded using $628 \ nm$ light,then the width of the slit is . . . . . . $\mu m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module