Analyse the observation table which shows variation of image distance [v] with object distance [u] in the case of a convex lens. Answer the questions that follow without doing any calculations.
`{:("Sl.No","Object distance","Image distance"),(,"u [cm]","v [cm]"),(1,-90,+18),(2,-60,+20),(3,-30,+30),(4,-20,+60),(5,-18,+90),(6,-10,+100):}`
(a) What is the focal length of the convex lens ? Given reasons to support your answer.
(b) Write the Sl.No of the observation which is not correct.
How did you arrive at this conciusion ?
(c ) Use an appropriate scale to draw the ray diagram for the observation at Sl. No 4 and find the approximate value of magnification.

Explanation

(a) When `u=-30 cm , v=+30 cm`. This indicates that the object is at C1 [which is `2F_(1)`]
r = 2f, 30 = 2f, f = 15 cm
Focal length of the convex lens is + 15 cm.
(b) If Object distance `u=-10 cm`, it means the object is placed between the optic centre and focus, a virtual erect image formed in this case on the same side as the object. Therefore v should be negative according to Cartesian sign convention ie.,
`v=-100 cm` and not `v=+100` cm as mentioned in the table.
Therefore SI. No 6 is incorrect.
(c ) SI. No 4 `u=-20 cm , v=+60 cm F_(1)=-15 cm 2F_(1)=-30 cm`
The object is between `F_(1)` and `2F_(1)`, an enlarged real and inverted image is formed beyond 2F,

Magnification `=m = (v)/(u)=((+60))/((-20))=-3`