Using monochromatic light of 410 nm wavelength,a single thin slit forms a diffraction pattern,with the first minimum at an angle of 40.0° from central maximum.The same slit,when illuminated by a new monochromatic light source,produces a diffraction pattern with the second minimum at a 60.0° angle from the central maximum.What is the wavelength of this new light?
A)276 nm
B)293 nm
C)309 nm
D)326 nm
E)342 nm

Question

Quizplus · Accepted Answer

The correct answer is A.The diffraction angle is given by:sin(theta) = (m * lambda) / dwhere:theta is the diffraction anglem is the order of the minimum (1 for the first minimum, 2 for the second minimum)lambda is the wavelength of the lightd is the width of the slitFor the first minimum, we have:sin(40.0°) = (1 * 410 nm) / dFor the second minimum, we have:sin(60.0°) = (2 * lambda) / dDividing the second equation by the first equation, we get:sin(60.0°) / sin(40.0°) = (2 * lambda) / (1 * 410 nm)Solving for lambda, we get:lambda = 276 nm

Using Monochromatic Light of 410 Nm Wavelength,a Single Thin Slit