The condition for an intensity minimum is b sin á = k ë
where b is the slit width.
For 474 nm, there are minima at
sin á = 6.21*10^-3 (k=1)
For 632 nm, there are minima at:
sin á = 8.283*10^-3 (k=1)
Clearly, the k=4 minimum of ë = 474 nm radiation coincides with the k = 3 minimum of 632 nm radiation. The distance from the central maximum is
d = 1.30 m*sin á = 3.23*10^-2 m
= 3.23 cm
For further reading, try
The a' symbol in my answer above was meant to be the diffraction angle alpha, and the e with two dots was supposed to be the wavelength, lambda
Let’s use the condition of diffraction minimum for one split of the width b
Since we have superposition of two maxima b•sinα is the same for two wavelengths,
λ1/λ2 = k2/k1,
632/474 = 4/3.
Therefore, k1=3, k2=4.
Now sinα = k1•λ1/b =3•632•10^-9/ 7.63•10^-5 =0.0248.
As the angle is very small tanα = sinα,
x=L• tanα =L• sinα =1.3•0.0248 =0.0323 m = 3.23 cm
Answer this Question
Physics - Light waves with two different wavelengths, 632 nm and 474 nm, pass ...
Physics help please! - Light waves with two different wavelengths, 632 nm and ...
physics - A flat screen is located 0.43 m away from a single slit. Light with a ...
physics - single-slit diffraction pattern is formed when light of 680.0 nm is ...
Physics - In a single-slit diffraction pattern on a flat screen, the central ...
Physics 2 - Light with λ = 530 nm passes through a single slit and then ...
Physics - Light of a red laser (wavelength λ=650 nm) goes through a narrow ...
Physics ~ important due tomorrow! - A student sets up a laser and screen and a ...
physics - In a double-slit experiment, two beams of coherent light traveling ...
Physics - A light beam ( 557.1 nm) illustrates a single slit 0.85mm wide. (a) ...