please give me full derivation involved in the interference of light waves and young's experiment.
Asked by priyadarshini_selvam  23rd Feb, 2012, 03:36: PM
The diagram below on the left depicts two sources labeled S_{1} and S_{2} and separated by some distance d. Point P is a point on the screen that happens to be located on some nodal or antinodal line; as such, there is an order value (m) associated with this point. Point C is the central point on the screen. The distance from point P to point C as measured perpendicular to the central antinodal line will be referred to as y. The screen is located a distance of L from the sources. In the following derivation, the wavelength of light will be related to the quantities d, m, y and L.
On the diagram above, source S_{2} is further from point P than source S_{1} is. The extra distance traveled by waves from S_{2} can be determined if a line is drawn from S_{1} perpendicular to the line segment S_{2}P. This line is drawn in the diagram on the left above; it intersects line segment S_{2}P at point B. If point P (a bright spot on the screen) is located a great distance from the sources, then it follows that the line segment S_{1}P is the same distance as BP. As such, the line segment S_{2}B is simply the path difference. That is, the small distance S_{2}B is equal to the difference in distance traveled by the two waves from their individual sources to point P on the screen. The logic is as follows:
Assertion
Logic/Rationale
i
S_{2}P = S_{2}B + BP
See diagram above left
ii
PS_{1}B = PBS_{1}
If screen is very far away (L >>> y), then lines S_{1}P and BP are 
iii
S_{1}P = BP
If S_{1}P and BP are  and line S_{1}B is perpendicular to BP, then the length BP = length S_{1}P
iv
S_{2}P = S_{2}B + S_{1}P
Since S_{1}P = BP, the expression S_{1}P can be substituted into i
v
S_{2}B = S_{2}P  S_{1}P
Algebraic manipulation of iv; subtract S_{1}P from each side
vi
S_{2}B is the PD
The path difference (PD) is defined as  S_{2}P  S_{1}P 
The yellow triangle in the diagram on the left above is enlarged and redrawn in the middle of the graphic. The triangle is a right triangle with an angle theta and a hypotenuse of d. Using the sine function, it can be stated that
sine() = S_{2}B / d
But since it has been previously stated that the path difference (PD) is equal to the length of the line segment S_{2}B, the above equation can be rewritten as
sine() = PD / d
It can be further asserted that the pink triangle ( S_{1}BS_{2}) and the yellow triangle ( ACP) in the diagrams above are similar triangles. To prove that any two triangles are similar, one must show that they have two corresponding angles that are equal. Since the line segment PC was drawn perpendicular to the central antinodal line, it forms a 90degree angle with the line AC. Thus, the corresponding angles S_{1}BS_{2} and ACP are equal.
The second set of corresponding angles of equal measure is S_{1}S_{2}B and APC. This can be proven by returning to the assumption that the screen is very far away (L >>> y). Clearly, the lines S_{1}S_{2} and PC are parallel lines. As such, the diagonal line S_{2}P creates two alternating angles that are equal in measure  that is, S_{2}PC = S_{1}S_{2}B. Since it is assumed that L >>> y, the line segment S_{2}P and AP are roughly parallel to each other and thus make roughly the same angle with the line segment S_{1}S_{2}. With this assumption, it is thus proven that APC = S_{2}PC. Since the S_{2}PC is equal to both S_{1}S_{2}B and APC, it follows that S_{1}S_{2}B = APC. As such, the pink triangle (S_{1}BS_{2}) and the yellow triangle (ACP) have two corresponding angles that are equal and thus are similar triangles.
If the pink triangle (S_{1}BS_{2}) and the yellow triangle (ACP) are similar, all corresponding angles are equal in measure, and so S_{2}S_{1}B = PAC. These two angles are labeled as theta in the diagrams above. The use of trigonometric functions allows one to relate the angle theta to the reliably measured distances d, y and L.
sine = PD / d
tangent = y / L
The above logic has consistently assumed that the screen upon which the interference pattern is projected is very far away; that is, L >>> y. This is typically the case for visible light interference patterns. In fact, the L value is typically on the order of several meters while the y value is on the order of a couple of centimeters. For such dimensions, the angle theta is less than 1 degree. For such small angles, the approximation that the sine = tangent can be made. Taking 1 degree as a sample angle, calculated values of the sine and tangent can be compared.
sine (1 deg) = 0.017452406
tangent (1 deg) = 0.017455064
Note that the values for the sine and the tangent of 1 degree show agreement out to the fourth significant digit. Since the sine and the tangent of these small angles are approximately equal, we can state that their ratio of lengths (as stated above) is also equal. That is,
PD / d = y / L
In the previous section of Lesson 3, it was shown that the path difference (PD) for any point on the pattern is equal to m , where m is the order number of that point and is the wavelength. By substitution,
m / d = y / L
As a final step in the derivation, the equation can be algebraically manipulated so that the wavelength () is by itself:
= y d / (m L)
The diagram below on the left depicts two sources labeled S_{1} and S_{2} and separated by some distance d. Point P is a point on the screen that happens to be located on some nodal or antinodal line; as such, there is an order value (m) associated with this point. Point C is the central point on the screen. The distance from point P to point C as measured perpendicular to the central antinodal line will be referred to as y. The screen is located a distance of L from the sources. In the following derivation, the wavelength of light will be related to the quantities d, m, y and L.
On the diagram above, source S_{2} is further from point P than source S_{1} is. The extra distance traveled by waves from S_{2} can be determined if a line is drawn from S_{1} perpendicular to the line segment S_{2}P. This line is drawn in the diagram on the left above; it intersects line segment S_{2}P at point B. If point P (a bright spot on the screen) is located a great distance from the sources, then it follows that the line segment S_{1}P is the same distance as BP. As such, the line segment S_{2}B is simply the path difference. That is, the small distance S_{2}B is equal to the difference in distance traveled by the two waves from their individual sources to point P on the screen. The logic is as follows:

Assertion  Logic/Rationale 
i  S_{2}P = S_{2}B + BP  See diagram above left 
ii  PS_{1}B = PBS_{1}  If screen is very far away (L >>> y), then lines S_{1}P and BP are  
iii  S_{1}P = BP  If S_{1}P and BP are  and line S_{1}B is perpendicular to BP, then the length BP = length S_{1}P 
iv  S_{2}P = S_{2}B + S_{1}P  Since S_{1}P = BP, the expression S_{1}P can be substituted into i 
v  S_{2}B = S_{2}P  S_{1}P  Algebraic manipulation of iv; subtract S_{1}P from each side 
vi  S_{2}B is the PD  The path difference (PD) is defined as  S_{2}P  S_{1}P  
The yellow triangle in the diagram on the left above is enlarged and redrawn in the middle of the graphic. The triangle is a right triangle with an angle theta and a hypotenuse of d. Using the sine function, it can be stated that
sine() = S_{2}B / dBut since it has been previously stated that the path difference (PD) is equal to the length of the line segment S_{2}B, the above equation can be rewritten as
sine() = PD / dIt can be further asserted that the pink triangle ( S_{1}BS_{2}) and the yellow triangle ( ACP) in the diagrams above are similar triangles. To prove that any two triangles are similar, one must show that they have two corresponding angles that are equal. Since the line segment PC was drawn perpendicular to the central antinodal line, it forms a 90degree angle with the line AC. Thus, the corresponding angles S_{1}BS_{2} and ACP are equal.
The second set of corresponding angles of equal measure is S_{1}S_{2}B and APC. This can be proven by returning to the assumption that the screen is very far away (L >>> y). Clearly, the lines S_{1}S_{2} and PC are parallel lines. As such, the diagonal line S_{2}P creates two alternating angles that are equal in measure  that is, S_{2}PC = S_{1}S_{2}B. Since it is assumed that L >>> y, the line segment S_{2}P and AP are roughly parallel to each other and thus make roughly the same angle with the line segment S_{1}S_{2}. With this assumption, it is thus proven that APC = S_{2}PC. Since the S_{2}PC is equal to both S_{1}S_{2}B and APC, it follows that S_{1}S_{2}B = APC. As such, the pink triangle (S_{1}BS_{2}) and the yellow triangle (ACP) have two corresponding angles that are equal and thus are similar triangles.
If the pink triangle (S_{1}BS_{2}) and the yellow triangle (ACP) are similar, all corresponding angles are equal in measure, and so S_{2}S_{1}B = PAC. These two angles are labeled as theta in the diagrams above. The use of trigonometric functions allows one to relate the angle theta to the reliably measured distances d, y and L.
sine = PD / d 

tangent = y / L 
The above logic has consistently assumed that the screen upon which the interference pattern is projected is very far away; that is, L >>> y. This is typically the case for visible light interference patterns. In fact, the L value is typically on the order of several meters while the y value is on the order of a couple of centimeters. For such dimensions, the angle theta is less than 1 degree. For such small angles, the approximation that the sine = tangent can be made. Taking 1 degree as a sample angle, calculated values of the sine and tangent can be compared.
sine (1 deg) = 0.017452406 

tangent (1 deg) = 0.017455064 
Note that the values for the sine and the tangent of 1 degree show agreement out to the fourth significant digit. Since the sine and the tangent of these small angles are approximately equal, we can state that their ratio of lengths (as stated above) is also equal. That is,
PD / d = y / LIn the previous section of Lesson 3, it was shown that the path difference (PD) for any point on the pattern is equal to m , where m is the order number of that point and is the wavelength. By substitution,
m / d = y / LAs a final step in the derivation, the equation can be algebraically manipulated so that the wavelength () is by itself:
= y d / (m L)
Answered by  23rd Feb, 2012, 10:47: PM
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