Question

Question 3 (2 points) This assignment is set up for sequential assessment. Complete each question and submit the answer before moving on to the next question. Correct answers for each question will be made available once the maximum number of attempts have been made for each submission. Refraction - CONVERGING and DIVERGING Lenses In these problems, you will solve for the object distance do, or focal length f, or the type of lens. Pay attention to the SIGN Convention. The sign of each quantity has meaning. CONVERGING Lens REAL focal points Positive focal length f Lens Equation do + = Magnification Equation hi ho m. DIVERGING Lens VIRTUAL focal points Negative focal length f Lens Equation do + d = } Magnification Equation = m. hod Note: case numbers don't necessarily match the situation numbers in the figures Case 1: The lens in a camera (which is a CONVERGING Lens) has a focal length f = 5.2 cm. If the distance between the lens and the film (where the image is formed, and it must be a REAL image) is 5.72 cm, how far away from the lens should the object be placed? Keep 2 decimal places Enter a number cm Submit (5 attempts remaining) CONVERGING Lens image formation 1 2 1477 1 REAL K-ff image 3 2 -d,

Answer 1

Case 1: Convexe lens Focal length, f = 5.2cm distance of image, d = 5.72cm distance of object do cambé calculated by lens equation ass ㅗ ㅗ do ㅗ di ca = | ) 1 ㅗ de f di 5.2. 5.72 ㅗ 0.192 - 0.175 =0.017 do do = 10.017 = 58.82 cm Hence, the object is placed 58.82 cm (Ans, behind the lens.

The image formed will be :- Case 2 As it uses magnifying glass which or 3 is a conven lens! Upright 2. Virtual Magnified langer than object option (g2 ks correct answers Also, distance of object --4.60 cm magnification, 3:0 (magnitude) = -3.0 Cactual) M = -di --3.0 do di = (3.0) do = (3.0) (-4.60) distance of image, Ida 13.80cm1 Answer Hence, Focal length, f cambe calculated as, f ㅗ f di do Lt-4 13.80 4.60) f = 6.89 cm Ans. We