Question 1 Ans is option E
Question 1 1 pts In the situation pictured below, the image forms: (A) To the left...
In the situation pictured below, the image forms: (A) To the left of F (B) Between F and the object (C) Between the object and the lens (D) Between the lens and F (right side) (E) To the right of F (right side) F F For best results: do NOT draw the object as large as in the previous question. AVOID putting the object close tof. In the situation pictured below, the image forms: (A) To the left of the...
Question 2 1 pts In this situation, the image forms: Draw the dashed lines behind the (A) To the left of C mirror (which indicate where it would look like the rays originate from if (B) Between C and the object viewed from the left side). Do this REGARDLESS of whether or not they (C) Between the object and f actually meet behind the mirror (D) Between fand the mirror (E) Behind the mirror (i.e., to the right of it)...
In this situation, the image forms:
(A) to the left of C
(B) Between f and the object
(C) Between the object and the mirror
(D) Behind the mirror (i.e., to the right of it)
с For best res as large as in puttir
In this situation, the image forms:
(A) To the left of C
(B) Between C and the object
(C) Between the object and f
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An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens. Part A What is the focal length of the lens? Part B Is the lens converging or diverging? Part C If the object is 8.00mm tall, how tall is the image? Part D Is it erect or inverted?
A converging lens with a focal length of 12.0cm forms a virtual image 8.00mm tall. Image is 17.0cm from the lens. Calculate (a)the position of the object, (b)magnification of the lens, (c)height of the object. (d)Is the image upright or inverted? (e) Are the object and image on the same side or opposite sides of the lens? (f) Draw ray-diagram
1 A converging lens with a focal length of 12.2 cm forms a virtual image 7.9mm tall, 11 2emto right of the lens. a. Determine the position of the object. b. Determine the size of the object. Is the image upright or inverted? Are the object and image on the same side or opposite sides of the lens? c. d. 2 You want to use a lens with a focal length of magnitude 36cm with the image twice as long...
A diverging lens with a focal length of -47.0 cm forms a virtual image 7.50 mm tall, 17.5 cm to the right of the lens.Part A Determine the position of the object. Part B Determine the size of the object.Part C Is the image erect or inverted? erect invertedPart D Are the object and image on the same side or opposite sides of the lens? The object and image are on the same side. The object and image are on...
Part A: A diverging lens has of focal length of 15.0 cm. An object is placed 21 cm to the left of the lens. a) draw a ray diagram showing the situation. b) find the location of the image produced by the lens (mind the signs). Part B: A converging lens is located 30 cm to the right of the previously mentioned diverging lens (part A). As a result, the image you found in part (a) is now instead located...
Constants An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens Part A What is the focal length of the lens? Tempistes Symbols Undo redo Pese keyboard shortcuts help Submit Request Answer Part B Is the lens converging or diverging? The lens is converging. The lens is diverging. Previous Answers ✓ Correct Part C If the object is 8.00 mm tall, how tall is the image?...