Question

The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation is ?y.
Part A What will happen to the fringe spacing if the wavelength of the light is decreased?

a. Δy will decrease
b. Δy will increase
c. Δy will not change

Part B What will happen to the fringe spacing if the spacing between the slits is decreased?

a. Δy will decrease
b. Δy will increase
c. Δy will not change

Part C What will happen to the fringe spacing if the distance to the screen is decreased?

a. Δy will decrease
b. Δy will increase
c. Δy will not change

Part D Suppose the wavelength of the light is 460 nm . How much farther is it from the dot on the screen in the center of fringe E to the left slit than it is from the dot to the right slit?

Answer 1

Answer:

Part A:

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

Part B:

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

Part C:

a) If the distance to the screen is decreased the fringe spacing will decrease.

Part D:

The dot in the center of fringe E is $920\ x\ 10^{-9} m$ farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

                      $\vartriangle y =\frac{m \lambda L}{d}$

                      $m=0,\pm 1,\pm 2,\pm 3,.....$

1. m is the order number.
2. $\lambda$ is the wavelength of the monochromatic light.
3. L is the distance between the screen and the two slits.
4. d is the distance between the slits.

• Part A:  a) In the above equation for the position of bright fringes we can see that if the wavelength of the light $\lambda$ is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.

• Part B:  b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
• Part C:  a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.

• Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at  the center of the fringe E in the screen we use the condition for constructive interference. That says that the  path length difference Δr between rays coming from the left and right slit must be $\vartriangle r=m \lambda$

        We simply replace the values in that equation :

                      $\vartriangle r= m \lambda =2.\ 460\ nm$

                      $\vartriangle r= 920\ x\ 10^{-9} m$

         The dot in the center of fringe E is $920\ x\ 10^{-9}m$ farther from the left slit than from the right slit.