A lens located in the y-z plane at x = 0 forms an image of an arrow at x = x₂ = 101.3 cm. The tip of the object arrow is located at (x,y) = (x₁, Y₁) = (-49.9 cm, 5.32 cm). The index of refraction of the lens is n = 1.49. (X,Yi).
A lens located in the y-z plane at x = 0 forms an image of an arrow at x = x₂ = 101.3 cm. The tip of the object arrow is located at (x,y) = (x₁, Y₁) = (-49.9 cm, 5.32 cm). The index of refraction of the lens is n = 1.49. (X,Yi).
Related questions
Question
Transcribed Image Text:A lens located in the y-z plane at x = 0 forms an image of an
arrow at x = x₂ = 101.3 cm. The tip of the object arrow is
located at (x,y) = (x₁, Y₁1) = (-49.9 cm, 5.32 cm). The index of
refraction of the lens is n = 1.49.
1) What is flens, the focal length of the lens? If the lens is converging flens is positive. It the lens is diverging, flens is
negative.
cm
Submit
2) What is y₂, the y-coordinate of the image of the tip of the arrow?
cm Submit
Virtual and upright
Virtual and inverted
(X₁,Y₁)
3) The lens is a plano-convex lens. What is Rens, the radius of curvature of the convex side of the lens?
cm Submit
X2
4) The object arrow is now moved to x = x1,new = -17.4 cm. What is X2,new, the new x-coordinate of the image of the
arrow?
cm Submit
5) Is the new image of the arrow real or virtual? Is it upright or inverted?
O Real and upright
Real and inverted
X
+
+
+
+
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
