for f, the focal 1 ti
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
Transcribed Image Text:4. Polystyrene with an index of refraction of 1.55 is used to create a thin lens. The front surface of the lens
is concave with a 30.0 cm radius of curvature. The back surface is convex with a 26.0 cm radius of
curvature. The image produced by the lens is 12.0 cm in front of the lens.
a) What is the value for f, the focal length, of the lens in em?
b) Determine the magnification produced by the lens.
c) A second lens with a focal length of +18.0 cm is placed 30.0 cm behind the first lens. Determine the
location of the final image, i2 , produced by the two-lens system in centimeters.
Expert Solution
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Step 1
Given:
The Polystyrene with an index of refraction (n)=1.55
The Radius of curvature (concave) =30 cm
The Radius of curvature (convex) = 26 cm
The image produced by the lens =12 cm
Step 2
a) On calculating the focal length (f) of the lens,
On substituting the given values in the above formula,
Therefore, the focal length of the image is 354.54 cm.
Step 3
b) On calculating the magnification of the lens as follows,
On substituting the given values in the above equation,
Step by step
Solved in 5 steps
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