An object 5 cm tall is placed 30 cm in front of a convex lens of focal length 20 cm.(a) Draw a ray diagram to show the formation of the image. 2(b) State the nature and position of the image. 1(c) Find the image distance and the height of the image.…

Answer: (b) Real, inverted and magnified, beyond 2F₂; (c) image distance = 60 cm on the other side of the lens; image height = 10 cm (inverted, i.e. −10 cm).

(a) Ray diagram to draw: Draw the principal axis and a convex lens at O. Mark F₁ and 2F₁ on the left and F₂ and 2F₂ on the right. Draw the object AB standing on the axis between F₁ and 2F₁.

• Ray 1: from A parallel to the principal axis; after refraction it passes through F₂.
• Ray 2: from A through the optical centre O; it goes undeviated.
• The two refracted rays meet below the axis beyond 2F₂ at A′. Draw A′B′ perpendicular to the axis.

Label: object AB, image A′B′, O, F₁, F₂, 2F₁, 2F₂, and arrows on all rays.

(b) Nature and position: The image is real, inverted and magnified, formed beyond 2F₂ on the other side of the lens.

(c) Using the Cartesian sign convention: u = −30 cm, f = +20 cm, h₁ = 5 cm.

\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \;\Rightarrow\; \frac{1}{v} = \frac{1}{20} + \frac{1}{-30} = \frac{3 - 2}{60} = \frac{1}{60} \;\Rightarrow\; v = +60\ \text{cm}\]

\[m = \frac{v}{u} = \frac{60}{-30} = -2, \qquad h_2 = m \times h_1 = -2 \times 5 = -10\ \text{cm}\]

The image is formed 60 cm from the lens on the other side; it is 10 cm tall, and the negative sign shows it is inverted (real). This agrees with (b): 60 cm is beyond 2F₂ (40 cm).