In a double-slit experiment,the wavelength la | vedclass

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

Question

(D) The condition for the $n^{th}$ interference maximum is $d \sin 	heta = n \lambda$. The condition for the first diffraction minimum is $a \sin 	h

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) The condition for the $n^{th}$ interference maximum is $d \sin 	heta = n \lambda$. The condition for the first diffraction minimum is $a \sin 	heta = \lambda$,which corresponds to the edge of the central diffraction envelope. The angular position of the first diffraction minimum is $\sin 	heta = \frac{\lambda}{a}$. Substituting this into the interference condition: $d \left( \frac{\lambda}{a} ight) = n \lambda$,which gives $n = \frac{d}{a}$. Given $d = 20 \ \mu m$ and $a = 4 \ \mu m$,we get $n = \frac{20}{4} = 5$. This means the $5^{th}$ interference maximum coincides with the first diffraction minimum. The central diffraction envelope contains the central maximum $(n=0)$ and the interference maxima on both sides up to $n=4$. The total number of bright fringes is $2n_{max} + 1 = 2(4) + 1 = 9$.

Similar Questions

When a thin transparent plate of thickness $t$ and refractive index $\mu$ is placed in the path of one of the two interfering waves of light,then the path difference changes by

In a double slit experiment,instead of taking slits of equal widths,one slit is made twice as wide as the other. Then in the interference pattern:

Vedclass Test Series

Exam Paper Generator

Online Exam Module