A coherent parallel beam of microwaves of wavelength `lambda = 0.5 mm` falls on aYoung's double- slit apparatus. The separation between the slits is 1.0 mm. The intensity of microwaves is measured on a screen placed parallel to the plane of the slits at a distance of 1.0 m from it as shown in Fig. 2.42.
If the incident beam makes an angle or `30^(@)` with the x-axis (as in the dotted arrow shown in the figure), find the y-coordinates of the first minima on either side of the central maximum.

Options

A. `(3)/(sqrt 7)` and `(1)/(sqrt 15) m`
B. `(3)/(sqrt 7)` and `(2)/(sqrt 15) m`
C. `(3)/(2 sqrt 7)` and `(1)/(sqrt 15) m`
D. `(6)/(sqrt 7)` and `(3)/(sqrt 15) m`

Answer

b

Explanation

Position of a point on a screen is
`y = D tan theta = 1 tan theta`
For first minima,
`n = 1, sin theta_(1) = (1)/(4), tan theta_(1) = (1)/(sqrt 15)`
For second minima,
`n = 2, sin theta_(2) = (3)/(4), tan theta_(2) = (3)/(sqrt 7)`
So, the positions of minima are
`y_(1) = tan theta_(1) = (1)/(sqrt 15) = 0.258 m`
`y_(2) = tan theta_(2) = (3)/(sqrt 7) = 1.13 m`
The minima are symmetrically placed on either of central mixima, therefore will be 4 minima at positions `+- 0.258 m` and `+- 1.13 m` on the screen.
a. When incident rays are incident normally, the waves arriving at slits are in phase, zero path difference before slit. Path difference after slits, at point P, is `d sin theta`.
Condition for minima at y-axis is
`d sin theta = (2 n - 1) (lambda)/(2)`
`sin theta = ((2 n -1)lambda)/(2d) = ((2 n - 1)(0.5))/(2 xx 1) = ((2 n - 1))/(4)`
As `sin theta le 1, ((2n - 1)/(4)) le` or `n le 2.5`
Hence, only first-order and second-order minima are possible.

b. Path difference before slits `= d sin phi`
Path difference after slits `= d sin theta`
As path of rays before slits is longer at `S_(1)` and `S_(2) P gt S_(1) P` after slits, so net path difference for first minima is
`d sin theta - d sin phi = +- (lambda)/(2)`
`sin theta = sin phi +- (lambda)/(2)`
`= sin 30^(@) +- (0.5)/(2 xx 1) = (3)/(4)` or `(1)/(4)`
`tan theta = (3)/(sqrt 7)` and `(1)/(sqrt 15)`
So, the position of first minima on either side of central maxima is
`y = D tan theta = (3)/(sqrt 7)` and `(1)/(15) m`