(8) A thin lens is made from material having an index of refraction of 1.65. The front surface of the lens has a radius of curvature of +12.0 cm. The back surface of the lens has a radius of curvature of -12.0 cm. (i) What is the focal length of the lens? (ii) Is this a converging or diverging lens? Justify your answer by showing your work. (iii) If an object is placed 18.0 cm in front of this lens, determine the location of the image formed (in cm). (iv) What is the magnification of this lens? (v) Is the image real or virtual? Justify your answer by showing your work.
(8) A thin lens is made from material having an index of refraction of 1.65. The front surface of the lens has a radius of curvature of +12.0 cm. The back surface of the lens has a radius of curvature of -12.0 cm. (i) What is the focal length of the lens? (ii) Is this a converging or diverging lens? Justify your answer by showing your work. (iii) If an object is placed 18.0 cm in front of this lens, determine the location of the image formed (in cm). (iv) What is the magnification of this lens? (v) Is the image real or virtual? Justify your answer by showing your work.
College Physics
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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On this question for v, it asked if it is a real or virtual image? How do i prove that?
![(8) A thin lens is made from material having an index of refraction of 1.65. The front surface of
the lens has a radius of curvature of +12.0 cm. The back surface of the lens has a radius of
curvature of -12.0 cm.
(i) What is the focal length of the lens?
(ii) Is this a converging or diverging lens? Justify your answer by showing your work.
(iii) If an object is placed 18.0 cm in front of this lens, determine the location of the image
formed (in cm).
(iv) What is the magnification of this lens?
(v) Is the image real or virtual? Justify your answer by showing your work.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0bff4613-fc62-4261-a036-97e7adfb62d1%2F9d5a4934-4ab1-4145-ae96-9b237b5dda61%2Frq3m53r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(8) A thin lens is made from material having an index of refraction of 1.65. The front surface of
the lens has a radius of curvature of +12.0 cm. The back surface of the lens has a radius of
curvature of -12.0 cm.
(i) What is the focal length of the lens?
(ii) Is this a converging or diverging lens? Justify your answer by showing your work.
(iii) If an object is placed 18.0 cm in front of this lens, determine the location of the image
formed (in cm).
(iv) What is the magnification of this lens?
(v) Is the image real or virtual? Justify your answer by showing your work.
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
Step 1
The focal length of the lens is calculated with lens makers formula .By substituting the radii of curvature in lens makers formula we get the focal length as
Given the object distance is u =18 cm
The lens equation is
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