Question

How did the focal length obtained from the intercept compare with that obtained from the distant object?

Slope of graph= -1             Focal length from intercept = 10 cm

Intercept of graph= 0.1     Focal length from distant object (find this value)

Table 2: Converging (Convex) Lens
Object distance, s (cm)	Image distance, s’ (cm)	1/s (cm)-1	1/s’ (cm)-1
12	45	0.08	0.02
15	31	0.07	0.03
20	20	0.05	0.05
25	17	0.04	0.06
30	15	0.03	0.07

Answer 1

(cm) 0.08 0.02 0.07 0.05 0.05 0:04 0.06 0.03 0.07 the from graph, intercept 3. Nalue of intercept: 41 divisions. 41 x 0.096 -0.012 10 = 0.0984 focal length = 10.16 cm 0.0984 distance object ie $ - $ = f NON को 는, value of focal length for distance object the is reciprocal of /s ak Same as intercept: enfato 16 cm. Ś =0 te this is

0.12 sett 0.096 0.072 I/s" o 0.048 0.024 0 0