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Assignment1_solution_update

# Assignment1_solution_update - something more will be needed...

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Assignment1 TA Solution Problem 1 4.6 m x 0 . 5 / 58 sin 0 = , x=4.24m 4.8 At the first mirror, i r θ θ = . For the second, i r i i θ θ θ - = - = 90 90 ' and ' ' i r θ θ = So i r θ θ - = 90 ' Problem 2 4.12 0 45 sin 1 sin 42 . 2 × = × θ so 0 17 = θ , the angular deviation is 45 0 -17 0 =28 0 4.17 θ sin 1 35 sin 33 . 1 0 × = × so 0 50 = θ 4.23 2 / sin sin 1 i i n θ θ × = × since ) 2 / cos( ) 2 / sin( 2 sin i i i θ θ θ = so ) 2 / cos( ) 2 / sin( 2 ) 2 / sin( 70 . 1 i i i θ θ θ = ; 0 6 . 63 = i θ 4.25 At the condition of “nearly perpendicularly watch”, the incident angle is very small, so θ θ θ tan sin m d d L d L 1278 . 1 5 . 1 33 . 1 tan 5 . 1 tan 33 . 1 sin 5 . 1 sin 33 . 1 2 2 1 2 1 2 1 = × = × × = × × = × θ θ θ θ similarly m d d L d d L 885 . 0 1 5 . 1 tan 1 tan 5 . 1 sin 1 sin 5 . 1 3 3 2 3 2 3 2 = × = + × × = × × = × θ θ θ θ n b =1.5 n w =1.33 n=1 θ 1 θ 2 θ 3

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Problem 3 5.10 ) / 1 / 1 /( ) 1 ( / 1 2 1 R R n f - - = where 1 2 R R - = so ) / 2 /( ) 1 ( / 1 1 R n f - = cm R 0 . 10 1 = cm s s f s i o i 1 . 1 0 . 1 / 1 10 / 1 / 1 / 1 / 1 - = - = - = 1 . 1 / = - = o i T s s M So the image is virtual, erect and larger than the object. Problem 4 5.29 m s s s s s M o o o o i T 8 . 0 6 . 0 / 1 3 / 1 / 1 ; / 3 = = + - = - = hence the distance is 0.8+0.8*3=3.2m Using large lens is to collect more intensity from the screen. The image will be inverted if it is to be real, so the set must be upside down or else
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Unformatted text preview: something more will be needed to flip the image. 5.30 for thin lens ) 1 1 ( 1 2 1 1 1 2 R R n n n f--= so in air ) 1 1 )( 1 ( 1 2 1 R R n f lm a--= ; in water ) 1 ( 1 3 / 4 ) 3 / 4 ( ) 1 1 ( 3 / 4 ) 3 / 4 ( 1 2 1-×-=--= lm a lm lm w n f n R R n f So a a lm lm w f f n n f 4 3 / 4 ) 1 ( 3 / 4 =--× = Problem 5 lens makers formula (1) use BK7 to produce a plano-convex lens, n600=1.516, R=f*(1.516-1)=31.0mm (2) use fused silica (or other material with good transmission in UV, visible and near infrared area) n350=1.477, R=-45*(1.477-1)=-21.5mm Problem 6...
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