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Unformatted text preview: Problem 1 Solution: From the geometry of the prism ( ) ( ) 1 1 2 2 i t t i δ θ θ θ θ = + 1 2 t i α θ θ = + Thus, 1 2 i t δ θ θ α = + When δ is smallest, 1 i d d δ θ = 2 1 1 1 t i i d d d d θ δ θ θ = + = , then 2 1 1 t i d d θ θ =  Using Snell’s Law, we have 2 2 sin sin t i n θ θ = , 1 1 sin sin i t n θ θ = Taking the derivative of Snell’s Law at each surface, we get 2 2 2 2 cos cos t t i i d n d θ θ θ θ = and 1 1 1 1 cos cos i i t t d n d θ θ θ θ = Note that from 1 2 t i α θ θ = + , we can get 1 2 1 t i d d θ θ =  Dividing the last two equations and substituting for the derivatives leads to 1 1 2 2 cos cos cos cos i t t i θ θ θ θ = Making use of Snell’s Law once again, 2 2 2 1 1 2 2 2 2 2 1 sin sin 1 sin sin i i t t n n θ θ θ θ = Since 1 n ≠ , it follows that 1 2 i t θ θ = , and therefore 1 2 t i θ θ = When δ is smallest, 1 2 m i δ α θ + = , 1 2 t α θ = Therefore, 1 1 sin sin 2 sin sin 2 m i t n δ α θ α θ + = = Problem 2 Solution (1) cornea (2) The closest distance for which the lens can accommodate to focus; near point is about 25 cm; far point is near infinity (3) retina (4) False. Cones are sensitive to color, the cones can be divided into ‘red’ cones, ‘green’ cones, and ‘blue’ cones based on measured response curves. They provide the eye’s color sensitivity. ELEC308 (Spring 2008) Midterm TA solution Problem 3 Solution a). Generally, the telephoto lens consists of a convex lens followed by a concave lens. b). The system matrix is + + =   = 1 180 1 180 1 180 180 1 1 1 1 1 180 1 1 1 1 2 1 2 1 2 1 1 2 f f f f f f f f M Since the effective focal length is 1000mm, and the back focal length is 50mm 1 2 1 2 1 180 1 1000 1 1 f f f f C f + =...
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This note was uploaded on 10/16/2011 for the course ELEC 308,315,10 taught by Professor Prof.shenghuisong during the Spring '11 term at CUHK.
 Spring '11
 Prof.ShenghuiSong

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