In YDSE distance between the S1 and S2 is d . P1 and P2 are two points equidistance from dotO at an angular position β as shown. A parallel beam of monochromatic light is incident at an angle α on the slits. Then the ratio of path difference at P1 and P2 is:

Question

In YDSE distance between the S1 and S2 is d . P1 and P2 are two points equidistance from dotO at an angular position β as shown. A parallel beam of monochromatic light is incident at an angle α on the slits. Then the ratio of path difference at P1 and P2 is: (A)  cot (α-β/2) cot (α+β/2) (B)  tan (α+β/2) cot (α-β/2) (C)  sin (α+β/2) cos (α-β/2) (D)  tan (α-β/2) cot (α+β/2)

Tardigrade Admin · Accepted Answer

(Δ P1/Δ P2)=( sin α- sin β/ sin α+ sin β)= tan (α-β/2) cot (α+β/2)

$cot \frac{α - β}{2} cot \frac{α + β}{2}$

$tan \frac{α + β}{2} cot \frac{α - β}{2}$

$sin \frac{α + β}{2} cos \frac{α - β}{2}$

$tan \frac{α - β}{2} cot \frac{α + β}{2}$

$\frac{Δ P _{1}}{Δ P _{2}} = \frac{s i n α - s i n β}{s i n α + s i n β} = tan \frac{α - β}{2} cot \frac{α + β}{2}$

Q. In YDSE distance between the S1​ and S2​ is d.P1​ and P2​ are two points equidistance from O˙ at an angular position β as shown. A parallel beam of monochromatic light is incident at an angle α on the slits. Then the ratio of path difference at P1​ and P2​ is:

cot2α−β​cot2α+β​

tan2α+β​cot2α−β​

sin2α+β​cos2α−β​

tan2α−β​cot2α+β​

ΔP2​ΔP1​​=sinα+sinβsinα−sinβ​=tan2α−β​cot2α+β​

$cot \frac{α - β}{2} cot \frac{α + β}{2}$

$tan \frac{α + β}{2} cot \frac{α - β}{2}$

$sin \frac{α + β}{2} cos \frac{α - β}{2}$

$tan \frac{α - β}{2} cot \frac{α + β}{2}$

$\frac{Δ P _{1}}{Δ P _{2}} = \frac{s i n α - s i n β}{s i n α + s i n β} = tan \frac{α - β}{2} cot \frac{α + β}{2}$