(a)(i) Define each of the following terms as it relates to converging lenses (i) focal length; (ii) optical Centre. (iii) Draw a ray diagram to illustrate h...

Assessment: WAEC SSCE - Physics - 2021 (Essay) Subject: Physics

Question 1 Report

(a)(i) Define each of the following terms as it relates to converging lenses (i) focal length; (ii) optical Centre.

(iii) Draw a ray diagram to illustrate how a converging lens is used to produce a virtual image of an object.

(b)(i) Name the primary colors of light. (ii) Match each primary color to its corresponding complementary color.

(c) A ray passes symmetrically through a glass prism of angle 60° and refractive index of 1.5. Calculate the angle of: (i) incidence; (ii) minimum deviation.

Answer Details

(a)(i) Focal length. The focal length of a converging lens is the distance from the optical centre of the lens to its principal focus (the point on the principal axis to which rays travelling parallel to the axis converge after refraction).

(a)(ii) Optical centre. The optical centre is the point at the middle of the lens through which a ray of light passes without being deviated (it travels straight on).

(a)(iii) Ray diagram: virtual image formed by a converging lens. When the object is placed between the lens and its principal focus (\(u < f\)) the lens acts as a magnifying glass: the emergent rays diverge and, produced backwards, meet on the same side as the object to form a virtual, erect and magnified image.

Two standard construction rays are used:

  • Ray 1 leaves the top of the object parallel to the principal axis and, after refraction, passes through the far principal focus \(F'\).
  • Ray 2 passes through the optical centre \(C\) of the lens and continues undeviated.

After the lens the two emergent rays diverge, so no real image is formed. Extending them backwards (dashed) they intersect on the same side as the object, locating the tip of the virtual image \(I\).

figure
Object between the converging lens and its focus F: emergent rays diverge and their backward extensions (dashed) meet at I, giving a virtual, erect, magnified image.

(b)(i) Primary colours of light. Red, Green and Blue.

(b)(ii) Complementary pairs. Each primary colour pairs with the colour obtained by mixing the other two primaries:

Primary colourComplementary colour
RedCyan
GreenMagenta
BlueYellow

(c) Ray passing symmetrically through a 60° prism, \(n = 1.5\). Symmetric passage means the ray traverses the prism at minimum deviation \(D_m\), so the refraction is described by

\[ n = \frac{\sin\!\left(\dfrac{A+D_m}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)}, \qquad A = 60^{\circ}. \]

(c)(i) Angle of incidence. At symmetric (minimum-deviation) passage the two refracting angles inside the prism are equal, each \(=\tfrac{A}{2}=30^{\circ}\), so the angle of incidence at the first face is

\[ n = \frac{\sin i}{\sin 30^{\circ}} \;\Rightarrow\; \sin i = 1.5 \times \sin 30^{\circ} = 1.5 \times 0.5 = 0.75, \] \[ i = \sin^{-1}(0.75) = 48.6^{\circ}. \]

(c)(ii) Angle of minimum deviation. Using the prism formula with \(i=\tfrac{A+D_m}{2}\):

\[ \sin\!\left(\frac{A+D_m}{2}\right) = n\sin\frac{A}{2} = 0.75 \;\Rightarrow\; \frac{A+D_m}{2} = 48.6^{\circ}, \] \[ A + D_m = 97.2^{\circ} \;\Rightarrow\; D_m = 97.2^{\circ} - 60^{\circ} = 37.2^{\circ}. \]

Angle of incidence \(i \approx 48.6^{\circ}\); angle of minimum deviation \(D_m \approx 37.2^{\circ}\).

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