(a) The variation of image distance (v) with object distance (u) for a convex lens is given in the following observation table. Analyse it and answer the questions that follow :

S. No.	Object distance (u) cm	Image distance (v) cm
1	-150	+30
2	-75	+37.5
3	-50	+50
4	-37.5	+75
5	-30	+150
6	-15	+37.5

(i) Without calculation, find the focal length of the convex lens. Justify your answer. (ii) Which observation is not correct? Why? Draw ray diagram to find the position of the image formed for this position of the object. (iii) Find the approximate value of magnification for $\mathrm{u} = - 30 \mathrm{cm}$. OR (b) (i) Define principal axis of a lens. Draw a ray diagram to show what happens when a ray of light parallel to the principal axis of a concave lens passes through it. (ii) The focal length of a concave lens is $20 \mathrm{cm}$. At what distance from the lens should a $5 \mathrm{cm}$ tall object be placed so that its image is formed at a distance of $15 \mathrm{cm}$ from the lens? Also calculate the size of the image formed.

(a) (i) The focal length of the convex lens is 30 cm. Justification: The focal length of a convex lens is the distance between the pole and the principal focus. From the given table, we can see that when the object distance is -150 cm, the image distance is +30 cm. This means that the image is formed at the principal focus, so the focal length is 30 cm. (ii) The observation that is not correct is the one with object distance -15 cm and image distance +37.5 cm. This is because the image distance should be less than the object distance for a convex lens. Ray diagram:  (iii) The approximate value of magnification for u = -30 cm is 5. Magnification = \frac{v}{u} = \frac{150}{-30} = -5 (b) (i) The principal axis of a lens is the line that passes through the center of the lens and is perpendicular to the lens. When a ray of light parallel to the principal axis of a concave lens passes through it, the ray is refracted away from the principal axis.  (ii) The object distance can be calculated using the lens formula: \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \frac{1}{-15} - \frac{1}{u} = \frac{1}{-20} \frac{1}{u} = \frac{1}{-15} + \frac{1}{-20} \frac{1}{u} = \frac{-4-3}{60} \frac{1}{u} = \frac{-7}{60} u = \frac{-60}{7} = -8.57 cm The size of the image can be calculated using the magnification formula: m = \frac{v}{u} = \frac{h'}{h} \frac{h'}{5} = \frac{-15}{-8.57} h' = \frac{-15 \times 5}{-8.57} = 8.75 cm