Using a converging lens of focal length 138.7 cm, we place an object at a distan
ID: 1373242 • Letter: U
Question
Using a converging lens of focal length 138.7 cm, we place an object at a distance 15.2 cm from the lens. How far away from the lens will the image appear?
A ray of light is incident from water with measured index of refraction 2.27 onto a translucent mineral. If the angle of incidence in the water is 26 degrees, and the angle of refraction is 36.91, degrees, what is the index of refraction of the mineral?
You have an object that is 56.6 cm from a lens, and the image appears 1.1 cm from the same lens. Calculate the magnification of the image.
Explanation / Answer
1.
From Thin Lens formula
1/f =1/di+1/do
=>1/di =1/f - 1/do =1/138.7 -1/15.2
di=-17.07 cm
2.
From snell's law
n1Sin(x1)=n2Sin(x2)
2.27*Sin(26) =n2*sin36.91
n2=1.657
3.
Magnifiaction
M=-di/do =-1.1/56.6
M=-0.01943
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