Question

038 (part 1 of 2) 10.0 points The second-order bright fringe (m= 2) is 4.37 cm from the center line.Determine time wavelength of the light. Answer in units of nm. 039 (part 2 of 2) 10.0 pointsCalculate the distance between adjacent bright fringes. Answer in units of cm.

Answer 1

we can use the small angle approximation so:

$\sin\theta \approx \tan \theta$

$d\sin\theta = m\lambda$ (maximum condition)

$d\frac{y}{L} = m\lambda$

so the wavelength is:

$\lambda = \frac{dy}{mL}=\frac{0.0284\times 10^{-3}(4.37\times 10^{-2})}{2(1.20)}=5.17\times 10^{-7} m = 517 nm$

the distance between adjacent bright fringes is:

$\Delta y = \frac{\lambda L}{d} = \frac{517\times 10^{-9}(1.20)}{0.0284\times 10^{-3}}=0.02184 = 2.184 cm$