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Q. 66

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
Found in: Page 958

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Short Answer

Light of wavelength passes though two slits separated by and is observed on a screen behind the slits. The location of the central maximum is marked on the screen and labeled

a. At what distance, on either side of , are the bright fringes?

b. A very thin piece of glass is then placed in one slit. Because light travels slower in glass than in air, the wave passing through the glass is delayed by in comparison to the wave going through the other slit. What fraction of the period of the light wave is this delay?

c. With the glass in place, what is the phase difference between the two waves as they leave the slits?

d. The glass causes the interference fringe pattern on the screen to shift sideways. Which way does the central maximum move (toward or away from the slit with the glass) and by how far?

(a) The distance of side is

(b) The fraction of the period of light wave is

(c) The phase between two waves is

(d) The central maximum move

See the step by step solution

Step by Step Solution

Step 1: Find the distance (part a)

We use the formula to location of the initial maximum, and we discover that

Step 2: Find fraction of period (part b)

(b) Keep in mind that the time of a wave with a speed of and a frequency of is . As a result, the ratio we're after is

In terms of numbers, we have

Step 3: Find phase between two waves (part c)

The aspect difference is simply our result amplified by

In terms of numbers, we have

Step 4: Explanation (part d)

The axis will be shifted so that the time it takes for the light from the two slits to reach it is the same for each. Because one of the laser sources will arrive later, the new base must be placed near to the slit with the glass.

Consider the two right triangles generated by the light beam when there is no glass (the hypotenuses are denoted by ) and when the glass is added (the hypotenuses are denoted by . It is clear that if we regard to be the quickest distance,

We may also infer from the Euclidean theorem that by examining this right triangle,

As a conclusion, our unknown will be

Lastly, it should be self-evident.

When we combine these, we obtain

Note that while we can give this statement in terms we know, it is much easier to calculate and then insert the values used in the aforementioned formula.

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