# What are the application of Newton's Ring

## Diffraction and interference

In order to calculate the length of the line \ (d \) as a function of the line lengths \ (r \) and \ (R \), we need some geometry from the intermediate level.

According to the hypotenuse of PYTHAGORAS in the yellow triangle, the following applies

\ [\ begin {eqnarray} {} R ^ 2 & = & (R - d) ^ 2 + r ^ 2 \ R ^ 2 & = & R ^ 2 - 2 \ cdot R \ cdot d + d ^ 2 + r ^ 2 \

0 & = & - 2 \ times R \ times d + d ^ 2 + r ^ 2 \ 2 \ times R \ times d - d ^ 2 & = & r ^ 2 \

d \ cdot (2 \ cdot R - d) & = & r ^ 2 \ end {eqnarray} \]**Comment:** The equation \ (r ^ 2 = d \ cdot (2 \ cdot R - d) \) also follows from the height theorem in the gray triangle with the height \ (r \).

Since \ (d \) can be neglected versus \ (2 \ cdot R \), one can set \ (2 \ cdot R - d \ approx 2 \ cdot R \). Hence \ [d = \ frac {r ^ 2} {2 R} \ qquad (3) \]

#### Determination of the wavelength

In the case of the reflected beam, the equations \ ((1) \) and \ ((3) \) for the path difference \ [\ Delta s = 2 \ cdot d + \ frac {\ lambda} {2} = \ frac {r ^ 2} {R} + \ frac {\ lambda} {2} \]

In the case of the continuous (transmitted) beam it follows from \ ((2) \) and \ ((3) \) \ [\ Delta s = 2 \ cdot d + \ lambda = \ frac {r ^ 2} {R} + \ lambda \]

To determine the wavelength of light, use is made of the fact that, for example, the path difference in the 2nd ring is \ (\ lambda \) less than the path difference in the 3rd ring. This results in the case of the transmitted beam

\ [\ lambda = \ Delta {s_3} - \ Delta {s_2} = \ left ({\ frac {{{r_3} ^ 2}} {R} + \ lambda} \ right) - \ left ({\ frac { {{r_2} ^ 2}} {R} + \ lambda} \ right) = \ frac {{{r_3} ^ 2 - {r_2} ^ 2}} {R} \] The corresponding calculation can also be used for the reflected Beam can be used.

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