A parallel beam of light of wavelength lambda | vedclass

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

Question

(B) The condition for the $n^{	ext{th}}$ diffraction minimum is given by the path difference $\Delta x = n\lambda$,where $n = 1, 2, 3, \dots$ is the

Vedclass · Accepted Answer

$4 \pi$

Vedclass · Answer

(B) The condition for the $n^{	ext{th}}$ diffraction minimum is given by the path difference $\Delta x = n\lambda$,where $n = 1, 2, 3, \dots$ is the order of the minimum.For the second minimum,we set $n = 2$,which gives the path difference $\Delta x = 2\lambda$.The relationship between phase difference $\phi$ and path difference $\Delta x$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x$.Substituting the value of $\Delta x$ for the second minimum:$\phi = \frac{2\pi}{\lambda} (2\lambda) = 4\pi$.Therefore,the phase difference between the rays coming from the two edges of the slit at the second minimum is $4\pi$.

Similar Questions

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 ....... $mm$.

$A$ diffraction pattern is formed by blue light. If the blue light is replaced by yellow light, then . . . . . . .

In a single slit diffraction experiment, the first minimum for red light $(660 \ nm)$ coincides with the first secondary maximum of some other wavelength $\lambda$. The value of $\lambda$ is $..... \mathring{A}$.

The width of the central maximum of a diffraction pattern on a single slit does not depend upon:

Vedclass Test Series

Exam Paper Generator

Online Exam Module