Question

Two thin lenses with a focal length of magnitude 21.0 cm, the first diverging and the second converging, are located 15.8 cm apart. An object 3.50 mm tall is placed 35.0 cm to the left of the first (diverging) lens.

1)How far from this first lens is the final image formed? Express your answer in centimeters.

2)Is the final image real or virtual?

3)What is the height of the final image? Express your answer in millimeters.

4)Is it upright or inverted?

Answer 1

15.8cm a 35um ist 2nd For the first lens ti -21, ų= -35 cm (on left) 80, using rene formula ti to 10-13 af 73 I 35 =- + 21 35) 5+3 105 105 V -13.125 cm 8 Now, This will act as for the Now, Total image object second lene Total distance of the object for lens. ų = -13.125+ 15.8) the converging ų -282925cm ta +21.cm by using lens Say lens formula ta VE I ų

(28.925) ta - ġ - 26.925 28.925-21 (28.925) (21) 7.925 607-425 Na - 76.646 cm of distance final Image = 15.8+ V from ist lene = 15.8+ 76.646 92.446 cm ane 2) The final Image formed on the Right the second lens So, It is Real Magnification due to first lens. M -13.125 - 35 0.375 Magnification due to second lens Image dutance 76.646 Object distance -28.925 > Image distance elect autane M2 - het Magneglication = m-M, M. - 2.649 (0-375)(-2-649) -0-993

Now, magnification, ma height of Image height of object height of mage -0.993 3.50mm Height of = Final Image -0.993) (3.50) - 3.47 mm ane m2 ) ve (4.) since, Magnification So, final Image is converted Ans