Question

Light of wavelength 550 nm falls on a

Answer 1

Answer:

The first diffraction maximum fringe will be at approximately 2.7 meters from the central maximum.

Explanation:

We can describe single slit diffraction phenomenon with the equation:

$a\sin\theta=m\lambda$ (1)

with θ the angular position of the minimum of order m respect the central maximum, a the slit width and λ the wavelength of the incident light. Because the distances between the first minima and the central maximum ($y_{m}$) are small compared to the distance between the screen and the slit (x), we can approximate $\sin\theta\approx\tan\theta=\frac{y}{x}$, using this on (1):

$a\frac{y_{m}}{x}=m\lambda$

solving for y

$y_{m}= \frac{mx\lambda}{a}$

Note that $y_{m}$is the distance between a minimum and the central maximum but we need the position of a maximum not a minimum, here we can use the fact that a maximum is approximately between two minima, so the first diffraction maximum fringe is between the minima of order 1 and 2, so we should find $y_{1}$, $y_{2}$ add them and divide by two:

$y_{1}= \frac{(1)(10.0m)(550\times10^{-9}\,m)}{3.00\times10^{-6}\,m}$

$y_{1}= 1.8 m$

$y_{2}= \frac{(2)(10.0m)(550\times10^{-9}\,m)}{3.00\times10^{-6}\,m}$

$y_{1}= 3.6 m$

$maximum = \frac{1.8+3.6}{2}=2.7m$