Phys_322_HW6

# Phys_322_HW6 - Phys 322 HW6 Solution 1 From Snell’s Law...

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Unformatted text preview: Phys 322 HW6 Solution 1. From Snell’s Law: n1 sin θ i = n2 sin θ t ; tan θ i = yo s o , tan θ t = − yi si ; For small θ , sin θ ≈ tan θ , thus 2. Eq. (5.16): y ns n1 yo ny = − 2 i , therefore M T = i = − 1 i yo n2 so so si 1 1 2 1 1 = (n − 1) , = (n − 1) − , where R2 = − R1 , so R R f R1 f 1 2 R1 = 2 f (n − 1) = 10cm . Eq. (5.17): 1 1 9 111111 10cm =−= − =− , so si = − +=, = −1.1cm . so si f si f so 10cm 1cm 10cm 9 Image is virtual, erect and larger than the object. 3. M T = −5 , so si = 5 ; and so + si = 60cm . The two equations can be easily so solved, then we have so = 10cm , si = 50cm . 50cm 111 1 1 6 = 8.3cm . =+= + = , so f = f so si 10cm 50cm 50cm 6 Similar as Problem 2, R1 = 2 f (n − 1) = 8.3cm 4. The image will be inverted if it’s to be real, so the set must be upside down or else something more will be needed to flip the image. M T = −3 , so si 1 1 1 1 + == . We can get so = 0.8m , hence the = 3 ; then s o 3 s o f 0 .6 m so distance is L = s o + si = 4so = 3.2m . 5. L = s o + si , so si = L − so ; and 111 1 1 1 + = , then + = . Getting rid of so si f so L − so f 2 the denominators, f (L − so ) + fs o = so (L − so ) , so − Ls o + Lf = 0 . 1 Solving the equation, so = d = s o + − so − = L ± L2 − 4 Lf ; so 2 L + L2 − 4 Lf L − L2 − 4 Lf − = L2 − 4 Lf ; 2 2 d 2 = L2 − 4 Lf , therefore f = L2 − d 2 4L 2 ...
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## This note was uploaded on 12/07/2011 for the course PHYS 322 taught by Professor Na during the Fall '11 term at Purdue.

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Phys_322_HW6 - Phys 322 HW6 Solution 1 From Snell’s Law...

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