1. A 1.7-cm-tall object is 14 cm in front of a-22 cm focal length. a. Calculate
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Question
1. A 1.7-cm-tall object is 14 cm in front of a-22 cm focal length. a. Calculate the image position b. Calculate the image height. 2. A 1.7-cm-tall object is 14 cm in front of a converging lens that has a 26 cm focal length. a. Calculate the image position b. Calculate the image height. 3. An object is 11 cm in front of a diverging lens with a focal length of -6 c. a. Determine the location of the image. 4. A 1.2-cm-tall object is 83 cm in front of a converging lens that has a 28 cm focal length. a. Calculate the image position b. Calculate the image height. 5. A 1.0-cm-thick layer of water stands on a horizontal slab of glass. A light ray in the air is incident on the water 55 o from the normal. a. What is the ray's direction of travel in the glass?Explanation / Answer
1) A) Given that u = 14 cm
f = -22 cm
apply 1/f = 1/u+1/v
-1/22 = 1/14+(1/v)
1/v = (-1/22)+(-1/14)
v = -8.55 cm
B) magnificatiion m = -v/u = h'/h
8.55/14 = h'/1.7
h' = 1.038 cm
2) u = 14 cm
f = 26 cm
v = ?
1/f = 1/u + 1/v
1/26 = 1/14 + 1/v
v = -30.33
B) h'/h = -v/u = 30.33/14
h' = 3.68 cm
3) f = -6 cm
u = 11 cm
-1/6 = 1/11 + 1/v
v = -3.8 cm
4) f = 28 cm
u = 83 cm
1/28 = 1/83 + 1/v
v = 42.25 cm
b) h'/h = -v/u = -42.25/83 = -0.509
h' = -0.509*1.2 = 0.61 cm
5) n1*sin(t1) = n2*sin(t2) for air water interface
1*sin(55) = 1.33*sin(t2)
sin(t2) = sin(55)/1.33 = 38 degrees
fro water and glass interface
n1*sin(38) = n2*sin(t3)
1.33*sin(38) = 1.5*sin(t3)
t3 = 33 degrees from the normal of the glass and water interface
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