Question

1) An object arrow 2 cm high is placed 20 cm from a converging lens of focal length of 10 cm. Find

1. The location of the final image

2. the nature of image,

3. the magnification and height of the image

4. Draw a ray diagram of the image formation     

2) Using the same lens as above, work out the results for an Object distance of 8 cm. Answer Parts (a) through (d) as above.

Answer 1

10cm Q] object distance = = -20cm focal length = f = Image distance av By using Lens formida I-I v u b ㅗ 20 = = I-I = 10 20 2-1 20 V= 20cm behind the ch ho = 20m Image is formed 20 height of object a ho height of image ch Magnification -m= = 20. Go) -2 m= h = -2= 2 h 2 => h = -4cm is 4cm & is inverted ROUGH Height of Image

Magnification M = m = V = u 20 -1 m = h. ho --/= h 2 => h = -2cm : Height of image is 2.cm & is invested

2F 2F -د I u V 2) Now object distance u= 8cm ㅗ V C 8 10 4-5 40 8 v= -40 cm from the lens same side as that i. Image is formed 40 cm and it is at the of object. Magnification = m = V = -40 lu -8 height of of image =h hox5 => h = 2x5=10cm height of im image = 10cm & is Sect. 2F 2 F tet F 2F ROLIGH