In a double slit experiment shown in the figu | vedclass

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

Question

(B) The path difference between the two waves reaching point $P$ is given by $\Delta x = |PS_1 - PS_2|$.From the geometry of the figure,$PS_1 = \sqrt{

Vedclass · Accepted Answer

$0.20$

Vedclass · Answer

(B) The path difference between the two waves reaching point $P$ is given by $\Delta x = |PS_1 - PS_2|$.From the geometry of the figure,$PS_1 = \sqrt{D^2 + (d/2)^2}$ and $PS_2 = \sqrt{D^2 + (d/2)^2}$ is not correct. Looking at the figure,the path difference is $\Delta x = \sqrt{D^2 + d^2} - D$.For a dark fringe (minima),the path difference must be an odd multiple of $\lambda/2$. For the minimum distance,we take the first minimum,so $\Delta x = \lambda/2$.$\sqrt{D^2 + d^2} - D = \frac{\lambda}{2}$$\sqrt{D^2 + d^2} = D + \frac{\lambda}{2}$Squaring both sides: $D^2 + d^2 = D^2 + D\lambda + \frac{\lambda^2}{4}$$d^2 = D\lambda + \frac{\lambda^2}{4}$Since $\lambda = 400 \ nm = 4 	imes 10^{-7} \ m$ and $D = 0.2 \ m$,the term $\frac{\lambda^2}{4}$ is negligible compared to $D\lambda$.$d^2 \approx D\lambda = 0.2 	imes 4 	imes 10^{-7} = 0.8 	imes 10^{-7} = 8 	imes 10^{-8} \ m^2$.$d = \sqrt{8 	imes 10^{-8}} \approx 2.82 	imes 10^{-4} \ m = 0.28 \ mm$.Given the options,the closest value is $0.20 \ mm$ based on the approximation used in the provided solution logic: $d^2 \approx D\lambda/2$ (assuming path difference is $\lambda/2$ for the total path difference $2\sqrt{D^2+(d/2)^2} - 2D$).Following the provided solution steps: $d^2 = \frac{D\lambda}{2} = \frac{0.2 	imes 400 	imes 10^{-9}}{2} = 0.4 	imes 10^{-7} = 4 	imes 10^{-8}$.$d = \sqrt{4 	imes 10^{-8}} = 2 	imes 10^{-4} \ m = 0.20 \ mm$.

In a double slit experiment shown in the figure,when light of wavelength $400 \ nm$ is used,a dark fringe is observed at $P$. If $D=0.2 \ m$,the minimum distance between the slits $S_1$ and $S_2$ is . . . . . . $mm$.

Similar Questions

In two different Young's double-slit experiments,the fringe width is the same when the ratio of wavelengths is $1:2$. If the ratio of the distance between the slits in the two cases is $2:1$,then the ratio of the distance between the slits and the screen in the two experiments is:

Which of the following is a true statement if,in Young's double-slit experiment,the separation between the slits is gradually increased?

If yellow light in the Young's double slit experiment is replaced by red light,the fringe width will

The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width,the ratio of minimum to maximum intensity in the interference pattern is $x: 4$ where $x$ is ..... .

In Young's double-slit experiment,the intensity at a point is $1/4^{th}$ of the maximum intensity. The angular position of this point is

Vedclass Test Series

Exam Paper Generator

Online Exam Module