HCI 10C Superposition Solutions (Self-Review Questions)

Hwa Chong Institution (College) C1 2023 1 Suggested Solutions Tutorial 10C: Single Slit and Rayleigh’s Criterion Self-Practice Questions S1 This question involves the use of the single slit diffraction formula sin b  = . Using sin b  = , 1 9 1 3 sin ( ) 589.0 10sin ( )0.250 10 0.135 b  − − − − = =  = Let w be the width of the central maximum and L be the slit-screen distance. Then, 0.5tan ( ) w L = 2 tan 2(2.50) tan(0.135 ) 0.0118 1.18 wL m cm = = = = Note: In this answer, we show how the width of the central maximum can be calculated exactly. This is in contrast to Lecture Example 10.7.1 in your notes, where we showed the use of small angle approximation to simplify our working. This question can also be solved in similar fashion by taking small angle approximation, which will still give you an answer of 1.18cm. Persuade yourself why this approximation is reasonable. Common Error: Remember that similar to the equation for the diffraction grating equation,  is taken with respect to the principle axis. central maximum, w Single-slit Diffraction of light L

Hwa Chong Institution (College) C1 2023 2 S2 This question involves the use of the single slit diffraction formula sin b  = . Using trigonometry, 20.5 0.5 3.8 10tan ( ) 0.726 1.5 w L −= =  =  The first minimum intensity occurs at angle  from the central maximum. sin b  = 3sin (0.050 10 ) sin(0.726 ) 633b nm −= =    = S3 This question involves qualitative understanding of the Rayleigh criterion. The Rayleigh criterion states that two images are just resolved or distinguishable when the central maximum of one image coincides with the first minimum of the other image. Ans: (D) S4 This question involves quantitative calculation using the Rayleigh criterion min b   . The Rayleigh criterion for the resolving power of a single aperture is given by min b   for a small min , where b is the diameter of the aperture and min is measured in radians. min 2 4 6.0 10 120 5.0 10 b rad  − − = = = i.e. 4 min 10 − Ans: (D) Note: This question is essentially the ‘reverse’ of Qu estion S1. Now, you are given the width of the central maximum (which effectively tells you the position of the first minimum intensity), from which you can use the formula to calculate a value for the unknown wavelength. Important: Remember to switch your calculator to RADIAN mode when solving calculation problems on resolution.