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David · Accepted Answer

Problem 5.13 
Answer
Given that, 
f1=  60mm 
f2=  24mm 
d = 42mm 
Equivalent Focal length in Huygen’s Eyepiece 
F = f1f2/f1+f2-d 
F = 60*24/60+24-42 
F = 34.28 mm (answer) 
Position of Second Principle Point in Huygen Eye piece is , 
β = f2d/f1+f2-d 
β = -24*42/24+60-42 
β = -24mm (From eye Lens) (Answer) 
Position of First Principle Point in Huygen Eye piece is , 
α = f1d/f1+f2-d 
α = 60*42/24+60-42 
α = +60mm (From field Lens) (Answer) 
Problem 5.16 
Answer
Focal length of a Focometer = 25cm 
Distance between the two graticule(On cross hairs S) = 8.75cm 
Size of the image (s’) = 6.75cm 
Let F is the focal length of the lens to be tested, 
Ffoco / Ftest lens = S/s’  
25cm/Ftest lens = 8.75/6.75 
Ftest lens = 17.857cm = 17.86cm ( rounding off hence 5 is increased by 1 and 7 is deleted) Answer
Problem 5.

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