(a)(i) State the laws of refraction of light.
(ii) Describe an experiment to determine the refractive index, n, of the material of an equilaleral triangular glass prism using the minimum deviation method.
(b) A rectangular glass prism of thickness 12cm is placed on a mark on a piece of paper resting on a horizontal bench:
(i) Draw a ray diagram to show the apparent position of the mark in the glass prism.
(ii) If the refractive-index of the material of the prism is 1.5, calculate the apparent displacement of the mark.
Explanation
(a)(i) The incident ray, the refracted ray, and the normal at the point of incidence, all lie in the same plane. OR The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant \(\frac{\text{Sin i}}{\text{Sin r}}\) = Constant
Place the triangular prism on the drawing sheet and trace the outline. Arrange two pins on one side vertically and using two other pins, determine the emergent ray on the other side. Remove the prism and. join the position of pins to meet the outline of the prism. Draw the normal at the point of incidence. Produce both the incident and the emergent rays to intersect and measure the angles of incidence IĀ° and the deviation dĀ°. Repeat the procedure for at least four other values of i. Plot a graph of d on the vertical axis against i on the horizontal axis. Obtain minimum deviation D as below.
n = \(\frac{sin \frac{1}{2}(A +D)}{sin \frac{1}{2} A}\)
= \(\frac{sin \frac{1}{2} (60 + D)}{sin 30}\)
Precautions
- neat traces
- pins fixed vertically
- Avoid parallax in reading protector
(b)(i)
(ii) n = \(\frac{\text{real depth}}{\text{apparent depth}}\)
apparent depth = \(\frac{\text{real depth}}{n}\)
= \(\frac{12}{1.5}\)
= 8cm
Apparent displacement = real depth - apparent depth
= 12 - 8 = 4cm