In Young's double-slit experiment,both slits | vedclass

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

Question

(C) In Young's double-slit experiment,the fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength

Vedclass · Accepted Answer

The pattern gets less bright and farther apart.

Vedclass · Answer

(C) In Young's double-slit experiment,the fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.As the viewing screen is moved farther from the slits,the distance $D$ increases.Since $\beta \propto D$,an increase in $D$ leads to an increase in the fringe width $\beta$,meaning the interference fringes move farther apart.Additionally,as the screen moves farther away,the intensity of light reaching the screen decreases due to the inverse square law and the spreading of the beam,making the pattern less bright.Therefore,the pattern becomes less bright and the fringes move farther apart.

In Young's double-slit experiment,both slits are illuminated by a laser beam and the interference pattern was observed on a screen. If the viewing screen is moved farther from the slit,what happens to the interference pattern?

Similar Questions

In a Young's double slit experiment,the distance between the two identical slits is $6.1$ times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single slit diffraction pattern is

Two sources of light of wavelengths $2500 \,\mathring{A}$ and $3500 \,\mathring{A}$ are used in Young's double slit experiment simultaneously. Which orders of fringes of the two wavelength patterns coincide?

For sustained interference fringes in a double-slit experiment,the essential condition$(s)$ is/are:
$1$. Sources must be coherent.
$2$. The intensities of the two sources must be equal.
Here,the correct option$(s)$ is/are:

In $YDSE$,the angular width of a fringe formed on a distant screen is $1^o$. The slit separation is $0.01\,mm$. The wavelength of light is:

If the slit widths of $YDSE$ are in the ratio $1:4$,then the ratio of maximum to minimum intensity on the screen is: (in $:1$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module