HCI 10B Superposition Solutions (Self-Review Questions)

Hwa Chong Institution (College) C1 2023 1 Suggested Solutions Tutorial 10B Young’s Double Slits and Diffraction Grating Self-Practice Questions on Young’s Double Slits S1 This question involves direct substitution of given quantities into the fringe separation equation for Young’s double slits. For Young’s double slits experiment, fringe separation d Ly = 6 4 3 (0.59 10 )(0.30) 4.92 100.36 10y − − − = =   m S2 This question involves the use of the fringe separation equation, similar to S1. Method 1 d Ly =  Distance of slits from screen, L = 0.34 m a. 6 3 (0.60 10 )(0.34) 0.510.60 10 Ly mmd  − −  = = =  b. 6 3 (0.70 10 )(0.34) 0.800.30 10 Ly mmd  − −  = = =  c. 6 3 (0.70 10 )(2 0.34) 1.600.30 10 Ly mmd  − −  = = =  Method 2 a. y    () 0.600.60( )0.70 0.51 new new old old yy mm   = = = b. 1y d   () 0.400.60( )0.30 0.80 old new old new dyy d mm  = = = c. Ly d   ( )( ) 0.400.60( )(2)0.30 1.60 old new new old new old dLyy dL mm  = = = Note: There are two ways to solve this question. In this question, since you are given explicit values for fringe separation, slit separation and wavelength, you can calculate explicitly the slit-screen distance L. A more general approach would be to consider the proportionalities of the quantities involved in the equation. This would be useful if explicit values were not given in the question.

Hwa Chong Institution (College) C1 2023 2 S3 This question is on fringe separation for Young’s Double Slits. By inspection, y is maximized when -  is maximized - L is maximized - d is minimized Ans: (C) Self-Practice Questions on Diffraction Grating S4 This question involves direct use of the diffraction grating equation. For diffraction grating, d sin  = n  31 x 10 sin(22.50 ) (1)( )600  − =   = 638 nm S5 This question involves direct use of the diffraction grating equation. For diffraction grating, d sin  = n   2)2(sin =d  )2(sin 2  =d Ans: (C) Note: d Ly = Question wants bright fringes to be farthest apart → they are asking what combination of ,,Ld would maximize the value of fringe separation y Reminder! Pay close attention to the units of N. In most cases, N is not given in SI units, which you need to account for when taking the reciprocal to find the slit separation d. Common Error: Remember that  in the diffraction grating equation refers to the angle with respect to the principle axis.

Hwa Chong Institution (College) C1 2023 3 S6 This question involves taking proportionalities of quantities in the diffraction grating equation. For diffraction grating, d sin  = n   sin for same d, n 9 sin()sin sin(23.7 )(435 10 )( ) sin(15.8 ) 643.3 red red blue blue nm   − = =  = Comparing to the data table, the impurity is cadmium. Ans: (B) S7 This question involves use of the diffraction grating equation, and subsequently calculating the maximum order of fringes observed. For diffraction grating, d sin  = n      = − d n 1sin Angular separation between the first order diffracted wavelengths, o 3 9 1 3 9 1 12 959.0 600/)10( )10x589)(1(sin 600/)10( )10x615)(1(sin =     −     =−= − − − − − − Max observable order for  = 589 nm max,519 3 9 sin(90 ) (10 / 600)(1) 589 10 2.8 2 nm dn  − − = =  = = Max observable order for  = 615 nm max,615 3 9 sin(90 ) (10 / 600)(1) 615 10 2.7 2 nm dn  − − = =  = = Note: Since n must be a whole number, the calculated values of n were rounded down. Note: Similar to S1, you can also calculate the slit separation explicitly for blue light, and subsequently use it to calculate the wavelength of the red light. This part is similar to Lecture Example 10.6 .2. Conceptually , it makes use of the fact that max 90 = for a diffraction grating experiment.

Hwa Chong Institution (College) C1 2023 4 S8 This question involves qualitative analysis of the diffraction grating equation. (i) d sin  = n sin ( ) n ndd  = = 2sin( ) 2sin( ) = 2 2 Ans: (C) Note: If you plot the graph of sin against  , you can observe that the angular displacement  between neighbouring orders increases with order n.) (ii) d sin  = n sin In other words,  increases with  , hence ' Ans: (C)