Post your solutions to the Mar 13 In-Class Exercise to this thread. Best, Chris

-- Mar 13 In-Class Exercise Thread

n1 = 1, n2 = 1.5 1) (n2 - n1)(1/r1 - 1/r2) = 0.25 1/r1 - 1/r2 = 0.25/0.5 1/r1 - 1/r2 = 0.5 Since |r1| = |r2|, r1 = 4, r2 = -4. ---- 2) (n2 - n1)(1/r1 - 1/r2) = -0.25 1/r1 - 1/r2 = -0.25/0.5 1/r1 - 1/r2 = -0.5 Since |r1| = |r2|, r1 = -4, r2 = 4

-- Mar 13 In-Class Exercise Thread

Assuming |r1| = |r2| (1.5 - 1)(1/r1 - 1/r2) = 1/4cm 1/2(1/r1 - 1/r2) = 1/4cm (1/r1 - 1/r2) = 2/4cm The value of r1 and r2 to have a focal length of 4cm would be 4 and -4 for a convex lens The value of r1 and r2 to have a focal length of -4cm would be -4 and 4 for a concave lens

-- Mar 13 In-Class Exercise Thread

Happy Wednesday, everyone! Here is my solution for the in-class exercise... If we assume n2 is the index of refraction for outside the lens, then n2 = 1 (air) If we assume n1 is the index of refraction for the inside of the lens, then n1 = 1.5 (glass) If r1 and r2 have the same magnitude then we can assume that r1 = -r2 Using the Lensmaker's equation: (n2 - n1)(1/r1 - 1/r2) = 1/f Find values of r1, r2 when f = 4cm (1.5 - 1)(2/r1) = 1/4 (.5)(2/r1) = 1/4 2/r1 = 1/2 r1 = 4 = -r2 r2 = -4 Find values of r1, r2 when f = -4cm (1.5 - 1)(2/r1) = 1/-4 (.5)(2/r1) = 1/-4 2/r1 = 1/-2 r1 = -4 = -r2 r2 = 4

-- Mar 13 In-Class Exercise Thread

<nowiki>For f = 4cm, n2 = 1.5 and n1 = 1 assuming r1 is the same magnitude of r2 (1.5 - 1)(1/r - 1/(-r)) = 1/4 (.5)(1/r + 1/r) = 1/4 (.5)(2/r) = 1/4 1/r = 1/4 1 = r/4 4 = r, so r1 = 4 & r2 = -4 For f = -4cm 1/r = 1/-4 1 = r/-4 -4 = r, so r1 = -4 & r2 = -4</nowiki>

-- Mar 13 In-Class Exercise Thread

For f = 4cm: We know @BT@|r_1| = |r_2|@BT@ and @BT@r_1 != r_2 => r_1 = -r_2@BT@ Let @BT@r_1@BT@ be @BT@r@BT@ @BT@1/2 (1/r - 1/r) = 1/4@BT@ @BT@1/2 (2 1/r) = 1/4@BT@ @BT@1/r = 1/4@BT@ @BT@=> r_1 = 4cm, r_2 = -4cm@BT@ For f = -4cm it is the opposite: @BT@r_1 = -4cm@BT@ and @BT@r_2 = 4cm@BT@

-- Mar 13 In-Class Exercise Thread

1. If f = 4, then (1.5-1)*(1/r1 - 1/r2) = 1/4 ==> r2 = -(2r1/(r1-2)). Which gives us the value r1 = 4 and r2 = -4. <br> 2. If f = -4, then (1.5-1)*(1/r1 - 1/r2) = -1/4 ==> r2 = 2r1/(r1+2). Which gives us the value r1 = -4 and r2 = 4.

-- Mar 13 In-Class Exercise Thread

r1, r2 have he same magnitude r1=-r2 f=4 (n2-n1)(1/r1-1/(r2)) = 1/4 (1.5-1)(1/r1-1/(r2)) = 1/4 (1.5-1)(1/r-1/(-r))=1/4 (0.5)(2/r) = 1/4 1/r=1/4 r=4 r1=4, r2=-4 f=-4 (1.5-1)(1/r1-1/(r2)) = -1/4 (1.5-1)(1/r-1/(-r))=-1/4 (0.5)(2/r) = -1/4 1/r=-1/4 r=-4 r1=-4, r2=4

-- Mar 13 In-Class Exercise Thread

n1 = 1, n2 = 1.5, |r1| = |r2| >> (n2 - n1)(1/r1 - 1/ r2) = 1/4 (1/2)(2/r1)=1/4 >> r1=r2=4