Question

A diverging lens has a focal length of magnitude 21.2 cm. (a) Locate the images for each of the following object distances. 42.4 cm distance cm location in front of A 21.2 cm distance location cm in front of 10.6 cm distance location in front of cm 4 (b) Is the image for the object at distance 42.4 real or virtual? o real o virtual Is the image for the object at distance 21.2 real or virtual? o real o virtual Is the image for the object at distance 10.6 real or virtual? o real o virtual (c) Is the image for the object at distance 42.4 upright or inverted? o upright inverted Is the image for the object at distance 21.2 upright or inverted? o upright o inverted Is the image for the object at distance 10.6 upright or inverted? o upright o inverted (d) Find the magnification for the object at distance 42.4 cm. Find the magnification for the object at distance 21.2 cm. Find the magnification for the object at distance 10.6 cm.

Answer 1

given that - 91.2 cm = 0.212 m P = 42.4cm 0.42um as given that f =-21.2cm po 42.4 cm from lens canation 3. *-- towy q - tichon - 14.133 cm location – infront of Image S -21-2 cm P - 21.2cm ŠŅba 5-1 -212 7-2 6) - q = - 10.6cm location: Infocont of the smage 8 - -21.2cm - 10.6 cm Ś. no 9.-7.066 em. location in front of someze

b) Image for the object at a distance ugyan is virtual a distance 21.2 is virtual a distance 10.6 is virtud magnification for us.uam is ma -9% ---14-133) 2.4 0-333 magnification for 21.2 cm is. me -9% - -1 -10.6) 21.2 0:5 magnification for 10.6.am is. man97 - -(-4.005) 10. = 0.666 magnification is (tve) so (at a distance 42.4 cm) Sorage is upright mogrification is ctive) so (at distance 21-2cm) Image is cupright magnification is five) so cot distance is labum) Srage is upright "