# Double Slit Help

1. Jun 10, 2005

### DDS

Consider the double-slit arrangement shown in the figure below, where the separation d is 0.305 mm and the distance L is 1.10 m.
A sheet of transparent plastic (n = 1.50) 0.0491 mm thick (about the thickness of a piece of paper) is placed over the upper slit. As a result, the central maximum of the interference pattern moves upward a distance y'. What is y'?

I know that Y=Ltan(angle) i have tried to manipulate the numbers and attempted to find theta but it just doesnt want to work....please any help is much appriciated.

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2. Jun 10, 2005

### DDS

can anyone help me please??

3. Jun 10, 2005

### apchemstudent

Was this the whole problem? No wavelength was given for the light shone through the slit?

4. Jun 10, 2005

### DDS

the problem is preventing me from solving the distance y, and i am asking if someone could help me do so

5. Jun 10, 2005

### apchemstudent

no i meant is this all the information that was given? Did you type out the question word for word or did you leave some information out?

6. Jun 10, 2005

### apchemstudent

There seems to be some information missing here. Either that, or we have to assume the difference in wavelength is 1...

7. Jun 10, 2005

### DDS

nope that the exact question word for word

8. Jun 11, 2005

### Staff: Mentor

All the needed information is given. The central maximum occurs where the phase difference between the two light waves equals zero. Introducing the plastic sheet adds a phase difference to the top wave.

Look at the picture. The bottom path is greater by $\Delta r$. Thus the phase shift that the top wave gets from passing through the plastic sheet must exactly equal the phase shift that the bottom wave gets by traveling the extra distance.

9. Jun 11, 2005

### DDS

So how exactly would i go about solving this.

if it be by calcuating the phase constant, could you give a hint as to how to start the calculations to determine it

10. Jun 11, 2005

### DDS

any hints as to where t oget me started....such as formulas or equations i can use or what to solve first??

11. Jun 11, 2005

### OlderDan

Before the plastic is inserted, the central maximum is at a point that is the same distance from both slits. After the plastic in inserted, the central maximum moves to a point where the "optical path" is the same from both slits. Optical path means number of wavelengths. From the information given for the central maximum, you can figure out the amount by which the physical path length of the light through the upper slit has been shortened by the plastic (more wavelengths will fit into the plastic than into the same thickness of air).

12. Jun 11, 2005

### DDS

can you elaborte on what i should a little more because i get the theory behind the problem but not how to apply..thats always been my problem with physics

13. Jun 12, 2005

### DDS

older dan could you possibly give me a verbal brake down as you did for the fluids questions??

much appriciated

14. Jun 12, 2005

### Staff: Mentor

As OlderDan explained, the "optical path" must be equal for both upper and lower waves. (That's the criteria for the formation of the central maximum.)

Now look at your diagram. The physical path of the lower waves is longer by an amount $\Delta r$. (You can express $\Delta r$ in terms of d, y, and L.) What is the increase in the optical path due to the light traveling the extra distance $\Delta r$? (How many wavelengths are contained in $\Delta r$?)

Now consider the upper wave. By inserting the plastic slab you have added wavelengths to the optical path. (The number of additional wavelengths must exactly equal the number of wavelengths in $\Delta r$, since the optical paths must be equal.) How many wavelengths? If the slab has thickness "t": How many wavelengths would that distance have had before the plastic was added? And how many wavelengths does it have after the slab is added? (Realize that the wavelength in plastic is smaller.) The difference is the additional optical path for the upper wave. Set that equal to what you found for the lower wave and solve for y.

15. Jun 12, 2005

### DDS

got it thanks

16. Jun 12, 2005

### OlderDan

I must say that I think this is a poor problem. Had it not been for the diagram, I would have said it is impossible to find a solution. If all you had to look at was the screen, you would have no way of locating what the diagam is calling the central maximum. There is a good chance that the postion of the "central maximum" is on or near a null of the single slit diffration pattern. There is no justification for assuming the single slit envelope pattern is going to be shifted to a new positon, and no justiication for rotating the plastic to be perpendicular to the direction from the slits to the point at y', although the effect of the latter will not be too great if the plastic is thin enough. You simply would not know how to locate the point where the optical paths are the same. It would depend on how many wavelengths fit into the plastic, and without knowing the wavelength that cannot be determined.

17. Jun 12, 2005

### DDS

Through a few formula manipulation i came up with thsi that gave me the asnwer, i approached it a very different way because i was confused by everyones suggestions.

y'=t(n-1)L / d

where t is the thickness of the plastic

18. Jun 12, 2005

### OlderDan

Looks good, as long as the small angle approximation (y' much smaller than L) is valid, which is a reasonable assumption unless you are told otherwise.