Calculating Distance Between Double Slits Using the Double-Slit Equation

[Doc Al]

Sorry one more question, m is equal to 2 yes?

Yes. In the formula for the single slit, the 2nd dark fringe corresponds to m = 2.

 

[Doc Al]

double slit problems

And here are solutions to some of the questions. Could you please just check my solutions.
1)Calculate the angle that will form between the centre maximum and the third-order dark fringe if the wavelength of the light that strikes two slits is 650 nm and the distance between the slits is 2.2 10-6 m.
1. n=3, λ=6.5* m, d=2.2* m
(n-0.5)λ= d*Sinθ
Sinθ= (n-0.5)*λ/d
= (3-0.5)6.5* /2.2*
= 2.5*6.5* /2.2
= 7.3863*0.1= 0.73863
Sinθ= 0.73863
θ= (0.73863)= 47.61 degrees

Careful: λ=650 nm (not 6.5 m); d = 2.2 10^-6 m (not 2.2 m).
Answer is correct.

2)In an experiment, blue light with a wavelength of 645 nm is shone through a double-slit and lands on a screen that is located 1.35 m from the slits. If the distance from the centre maximum to the 8th order bright fringe is 2.6 cm, calculate the distance between the two slits.

2. m=8, λ=6.45* m, L =1.35m, X=0.026m
X/L=mλ/d
d=Lmλ/X
d= 1.35*6.45* *8/0.026
=2.68* m

Correct formula, but the wavelength is 645 nm, not 6.45 m.
Answer is off by several factors of 10.

3)In an experiment, the distance from one slit to the third dark fringe is found to be 2.200 046 8 m. If the wavelength of the light being shone through the two slits is 590 nm, calculate the distance from the second slit to this same dark fringe.
3. n=3, PS1= 2.2000468m,λ=5.9* m
(n-0.5)λ=lPS1-PS2l
+/- (3-0.5)5.9* = 2.2000468-PS2
+/- 14.75* = 2.2000468-PS2
PS2= 2.2000468-14.75* OR 2.2000468+14.75*
PS2= 2.200045325m or 2.200048275m

I can't follow what you're doing here. What must be the difference in path length from slit to screen for the light from each slit to produce a dark fringe?

4)Two slits are separated by a distance of 2.00 10-5 m. They are illuminated by light of wavelength 5.60 10-7 m. If the distance from the slits to the screen is 6.00 m, what is the separation between the central bright fringe and the third dark fringe?

4. X/L= (2n+1)λ/2d
X= (2n+1)λL/2d= 42*5.60*10-6/4*10-5
= 58.8*10-2= 0.60m

Your formula is incorrect; compare with the correct formula that you used for problem 1. (You found the position of the 4th dark fringe, not the 3rd.)

 

[Doc Al]

single slit problems

Light of wavelength 625 nm shines through a single slit of width 0.32 mm and forms a diffraction pattern on a flat screen located 8.0 m away. Determine the distance between the middle of the central bright fringe and the first dark fringe

λ=6.25*〖10〗^(-7)m, w=0.32*〖10〗^(-3)m, L=8.0m
λ/w=X/L
X=Lλ/w= 8*6.25*〖10〗^(-7)/0.32*〖10〗^(-3)
= 156.25*10000
1.56*〖10〗^6m

Correct formula, but redo the arithmetic. (Keep track of those exponents.)

Light of 600.0 nm is incident on a single slit of width 6.5 mm. The resulting diffraction pattern is observed on a nearby screen and has a central maximum of width 3.5 m. What is the distance between the screen and the slit?

λ=6.00*〖10〗^(-7)m, w=6.5*〖10〗^(-3)m, 2X=3.5;X=1.75m
λ=wX/L
L=wX/λ= 6.5*1.75*〖10〗^(-3)/6*〖10〗^(-7)= 11.375*〖10〗^4/6 =2.00*〖10〗^4m

For some reason, you rounded off your answer. Your answer is OK given the data quoted, but I suspect that the central maximum width is not 3.5 m (more like 3.5 mm).

A monochromatic beam of microwaves with a wavelength of 0.052 m is directed at a rectangular opening of width 0.35 m. The resulting diffraction pattern is measured along a wall 8.0 m from the opening. What is the distance between the first- and second-order dark fringes?

. λ=0.052m, w=0.35m, L=8m
λ/w=X/L
X= Lλ/w= 8*0.052/0.35
= 1.18 m

OK.

Light from a red laser passes through a single slit to form a diffraction pattern. If the width of the slit is increased by a factor of two, what happens to the width of the central maximum?

λ1=wy1/L, y1= Lλ/w
λ2=2wy/L, y2= Lλ/2w
y2/y1= λ2L*w/2w*Lλ1
y2/y1= 0.5
y2=0.5y1

Good. The width of the central maximum decreases by a factor of two.

 

[pinkyjoshi65]

umm..ok so for Q2 above, the asnwer is 2.67*10^-4 m. For question3 I used the formula (n-0.5)=wavelength= lPS1-PS2l (absolute value), where PS1 is the distance of the 1st slit to the fringe and PS2 is the distance of the 2nd slit to the same fringe. For Q4, this was the formula that rhbhat had given me. I will use (n-0.5)wave= dX/L for this question

 

[Doc Al]

umm..ok so for Q2 above, the asnwer is 2.67*10^-4 m.

Good.

For question3 I used the formula (n-0.5)=wavelength= lPS1-PS2l (absolute value), where PS1 is the distance of the 1st slit to the fringe and PS2 is the distance of the 2nd slit to the same fringe.

Now I see. Good.

For Q4, this was the formula that rhbhat had given me. I will use (n-0.5)wave= dX/L for this question

Good. That's the one you need.

 

[pinkyjoshi65]

Thankz a lot Doc Al! Really really appreciate it..:)