(a) Define convex lens and give its nature of refraction.

(b) A 6 cm tall object is placed perpendicular to the principal axis of convex lens of focal length 25 cm. The distance of the object from the lens is 40 cm.

By calculation determine

1. the position and

2. the size of the image formed.

(a) A lens which is thicker at the centre and thinner at its ends is called convex lens. It converges the light rays passing through and is called a converging lens.

(b) Given,

f = 25 cm

u = -40 cm

v = ?

As we know, the lens formula is —

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

Substituting the values in the formula, we get,

$\dfrac{1}{25} = \dfrac{1}{v} - \dfrac{1}{-40} \\[1em] \dfrac{1}{25} = \dfrac{1}{v} + \dfrac{1}{40} \\[1em] \dfrac{1}{v} = \dfrac{1}{25} - \dfrac{1}{40}\\[1em] \dfrac{1}{v} = \dfrac{8-5}{200} \\[1em] \dfrac{1}{v} = \dfrac{3}{200} \\[1em] \Rightarrow v = \dfrac{200}{3} = 66.67 \text{ cm}$

∴ Image distance is 66.67 cm

u = -40 cm

v = 66.67 cm

height of object (h_o) = 6 cm

height of image (h_i) = ?

From formula:

$m = \dfrac{v}{u} = \dfrac{h$i}{ho}

Substituting the values in the formula, we get,

$\dfrac{66.67}{-40} = \dfrac{\text{h}$i}{6} \\[0.5em] \text{h}i = 6 \times -\dfrac{6.67}{40} \\[0.5em] \text{h}_i = -10 \text{ cm} \\[0.5em]

Hence, height of image = -10 cm